| Exercise
7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
Exercise 7.3
Question 1
Determine whether x=\frac{3}{2} and x=-\frac{4}{3} are the solutions of the equation 6x2-x-12=0 or not.
Sol :Put both the values of x in the equation.
When x=\frac{3}{2}
6\left(\frac{3}{2}\right)^{2}-\frac{3}{2}-12=0
6 \times \frac{9}{4}-\frac{3}{2}-12=0
\frac{54-48-6}{4}
\frac{54-54}{4}
= 0
When x=-\frac{4}{3}
6\left(-\frac{4}{3}\right)^{2}-\frac{4}{3}-12=0
6\left(\frac{16}{9}\right)+\frac{4}{3}-12=0
\frac{96+12-108}{9}
\frac{108-108}{9}
= 0
R.H.S = L.H.S
Therefore, \mathrm{x}=\frac{3}{2} and x=-\frac{4}{3} are the solutions of the given equation.
Put both the values of x in the equation.
When x = 1
12 – 5(1) + 4
1 – 5 + 4
= 0
Therefore, it is the solution to the equation.
When x = 3
32 – 5(3) + 4 = 0
9 – 15 + 4
= –2
Therefore, it is not the solution to the equation.
Put both the values of x in the equation.
When x = √3
(√3)2 –3√3(√3) + 6 = 0
3 – (3)3 + 6
3 – 9 + 6
9 – 9
= 0
Therefore, it is the solution to the equation.
When x = –2√3
(–2√3)2 –3√3(–2√3) + 6 = 0
12 + 18 + 6
= 36
Therefore, it is not the solution to the equation.
6\left(-\frac{4}{3}\right)^{2}-\frac{4}{3}-12=0
6\left(\frac{16}{9}\right)+\frac{4}{3}-12=0
\frac{96+12-108}{9}
\frac{108-108}{9}
= 0
R.H.S = L.H.S
Therefore, \mathrm{x}=\frac{3}{2} and x=-\frac{4}{3} are the solutions of the given equation.
Question 2
Determine whether (i) x = 1, (ii) x = 3 are the solutions of the equation x2 — 5x + 4 = 0 or
not.
Sol :Put both the values of x in the equation.
When x = 1
12 – 5(1) + 4
1 – 5 + 4
= 0
Therefore, it is the solution to the equation.
When x = 3
32 – 5(3) + 4 = 0
9 – 15 + 4
= –2
Therefore, it is not the solution to the equation.
Question 3
Determine whether x = √3 and x = —2√3 are solutions of the equation x2 – 3√3x + 6 =
0
Sol :Put both the values of x in the equation.
When x = √3
(√3)2 –3√3(√3) + 6 = 0
3 – (3)3 + 6
3 – 9 + 6
9 – 9
= 0
Therefore, it is the solution to the equation.
When x = –2√3
(–2√3)2 –3√3(–2√3) + 6 = 0
12 + 18 + 6
= 36
Therefore, it is not the solution to the equation.
Question 4
For 2x2— 5x —3 = 0, determine which of the following are solutions?
(i) x = 3
(ii) x= –2
(iii) x=-\frac{1}{2}
(iii) x=-\frac{1}{2}
(iv) x=-\frac{1}{3}
Sol :
The two possible solutions are x = 3 and =-\frac{1}{2}
Sol :
The two possible solutions are x = 3 and =-\frac{1}{2}
Since this question is given in standard form,
meaning that it follows the form: ax2 + by + c = 0, we can use the quadratic formula to solve for
x:
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}
x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(2)(-3)}}{2 \times 2}
x=\frac{5 \pm \sqrt{25+24}}{4}
x=\frac{5 \pm 7}{4}
x=\frac{5+7}{4}, x=\frac{5-7}{4}
\mathrm{x}=\frac{12}{4}, \mathrm{x}=\frac{-2}{4}
x = 3 , x=-\frac{1}{2}
That value of x is correct as well!
Therefore, the two possible solutions are:
x=3
x=−0.50
Put x = √2
(√2)2 + √2(√2) – 4 = 0
2 + 2 – 4
4 – 4
= 0
Therefore, it is the solution to the equation.
When x = –2√2
(–2√2)2 + √2(–2√2) – 4 = 0
8 – 4 – 4
= 0
Therefore, it is the solution to the equation.
Put x = –3 in the equation.
(–3)2 + 6(–3) + 9
9 – 18 + 9
=0
Hence it is a solution
Put x = –3 in the equation.
2(–3)2 + 5(–3) –3 = 0
18 – 15 –3
18 – 18
= 0
Therefore, x = –3 is the solution of the equation.
The given quadratic equation is 3x2 + 13x + 14 = 0
Putting x = – 2,
L.H.S.
3.(–2)2 + 13.(–2) + 14
3 x 4 – 26 + 14
12 – 26 + 14
26 – 26
= 0
Hence, x = – 2 is a solution of 3x + 13x + 14 = 0
Sol
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}
x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(2)(-3)}}{2 \times 2}
x=\frac{5 \pm \sqrt{25+24}}{4}
x=\frac{5 \pm 7}{4}
x=\frac{5+7}{4}, x=\frac{5-7}{4}
\mathrm{x}=\frac{12}{4}, \mathrm{x}=\frac{-2}{4}
x = 3 , x=-\frac{1}{2}
That value of x is correct as well!
Therefore, the two possible solutions are:
x=3
x=−0.50
Question 5
Determine whether (i) x= √2, (ii) x = –2√2 are the solutions of the equation x2 + √2 x – 4 =
0 or not.
Sol :Put x = √2
(√2)2 + √2(√2) – 4 = 0
2 + 2 – 4
4 – 4
= 0
Therefore, it is the solution to the equation.
When x = –2√2
(–2√2)2 + √2(–2√2) – 4 = 0
8 – 4 – 4
= 0
Therefore, it is the solution to the equation.
Question 6
Show that x = — 3 is a solution of x2 + 6x + 9 = 0.
Sol :Put x = –3 in the equation.
(–3)2 + 6(–3) + 9
9 – 18 + 9
=0
Hence it is a solution
Question 7
Show that x = — 3 is a solution of 2x2 + 5x –3 = 0.
Sol :Put x = –3 in the equation.
2(–3)2 + 5(–3) –3 = 0
18 – 15 –3
18 – 18
= 0
Therefore, x = –3 is the solution of the equation.
Question 8
Show that x = — 2 is a solution of 3x2 + 13x + 14 = 0.
Sol :The given quadratic equation is 3x2 + 13x + 14 = 0
Putting x = – 2,
L.H.S.
3.(–2)2 + 13.(–2) + 14
3 x 4 – 26 + 14
12 – 26 + 14
26 – 26
= 0
Hence, x = – 2 is a solution of 3x + 13x + 14 = 0
Question 9
For what value of k, x=\frac{2}{3} is the solution of the
equation
kx2 – x – 2 = 0.Sol
x=\frac{2}{3}
kx2 – x – 2 = 0
k\left(\frac{2}{3}\right)^{2}-\frac{2}{3}-2
\frac{4 \mathrm{k}}{9}-\frac{2}{3}-2
\frac{4 \mathrm{k}-6-18}{9}
4k – 24 = 0
4k = 24
k = 6
Put the value of x in the equation.
3x2 + 2kx — 3 = 0
3\left(-\frac{1}{2}\right)^{2}+2 \mathrm{k}\left(-\frac{1}{2}\right)-3=0
\frac{3}{4}-\frac{2 \mathrm{k}}{2}-3=0
\frac{3-4 k-12}{4}=0
–9 – 4k = 0
–4k = 9
\mathrm{k}=-\frac{9}{4}
Put x = 3/4
ax2 + bx — 6 = 0
a\left(\frac{3}{4}\right)^{2}+b\left(\frac{3}{4}\right)-6=0
\frac{9 a}{16}+\frac{3 b}{4}-5=0
\frac{9 a+12 b-96}{16}=0
9a + 12b – 96 = 0 divide by 3
3a + 4b – 32 = 0
3a + 4b = 32 (1)
kx2 – x – 2 = 0
k\left(\frac{2}{3}\right)^{2}-\frac{2}{3}-2
\frac{4 \mathrm{k}}{9}-\frac{2}{3}-2
\frac{4 \mathrm{k}-6-18}{9}
4k – 24 = 0
4k = 24
k = 6
Question 10
For what value of k, \mathrm{x}=-\frac{1}{2} is a solution of the equation
3x2 + 2kx — 3 = 0
Sol :Put the value of x in the equation.
3x2 + 2kx — 3 = 0
3\left(-\frac{1}{2}\right)^{2}+2 \mathrm{k}\left(-\frac{1}{2}\right)-3=0
\frac{3}{4}-\frac{2 \mathrm{k}}{2}-3=0
\frac{3-4 k-12}{4}=0
–9 – 4k = 0
–4k = 9
\mathrm{k}=-\frac{9}{4}
Question 11
For what values of a and b, x=\frac{3}{4} and x = — 2 are solutions of the
equation ax2 + bx — 6 = 0.
Sol :Put x = 3/4
ax2 + bx — 6 = 0
a\left(\frac{3}{4}\right)^{2}+b\left(\frac{3}{4}\right)-6=0
\frac{9 a}{16}+\frac{3 b}{4}-5=0
\frac{9 a+12 b-96}{16}=0
9a + 12b – 96 = 0 divide by 3
3a + 4b – 32 = 0
3a + 4b = 32 (1)
Put x = –2
ax2 + bx — 6 = 0
a(–2)2 + b(–2) – 6 = 0
4a – 2b – 6 = 0
4a – 2b = 6 (2)
ax2 + bx — 6 = 0
a(–2)2 + b(–2) – 6 = 0
4a – 2b – 6 = 0
4a – 2b = 6 (2)
Eliminate (1) and (2)
3a + 4b = 32
4a – 2b = 6 ×2
3a + 4b = 32
4a – 2b = 6 ×2
\begin{aligned}3a+4b&=32\\8a-4b&=12\\ \hline
11a&=44\end{aligned}
a = 4
Put a = 4 in equation (1).
3a + 4b = 32
3(4) + 4b = 32
12 + 4b = 32
4b = 32 – 12
4b = 20
b = 5
Question 12
For what value of k, x = a is a solution of the equation
x2 – (a + b) x + k = 0.Sol :
x2 – (a + b)x + k = 0
Put x = a
a2 – (a + b) a + k
a2 – a2 + ab + k = 0
ab + k = 0
k = –ab
Question 13
Determine the value of k, a and b in each of the following quadratic equation, for which the given value of
x is the root of the given quadratic equation:
(i) kx2 — 5x + 6 = 0 ; x = 2(ii) 6x2 + kx — √6 = 0; x=-\frac{\sqrt{3}}{2}
(iii) ax2 — 13x + b = 0; x =2 and x = —2 find a, b
(iv) ax2+ bx – 10 = 0; x=-\frac{2}{5} and x=\frac{5}{3}
Sol :
(i) kx2 – 5x + 6 = 0
Put x = 2
(ii) k2 – 5(2) + 6 = 0
4k – 10 + 6 = 0
4k = 4
k = 1
(i) kx2 – 5x + 6 = 0
Put x = 2
(ii) k2 – 5(2) + 6 = 0
4k – 10 + 6 = 0
4k = 4
k = 1
(iii) 6x2 + kx -√6 = 0
Put x = – 2/√3
6\left(-\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{k}\left(-\frac{\sqrt{3}}{2}\right)-\sqrt{6}=0
Put x = – 2/√3
6\left(-\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{k}\left(-\frac{\sqrt{3}}{2}\right)-\sqrt{6}=0
\frac{18}{4}-\frac{\sqrt{3}
\mathrm{k}}{2}-\sqrt{6}=0
\frac{18-2 \sqrt{3} \mathrm{k}-4 \sqrt{6}}{4}=0
18 – 2√3 k – 4√6 = 0
18 – 4√6 = 2√3k
\frac{18-4 \sqrt{6}}{2 \sqrt{3}}=\mathrm{k}
\frac{(18-4 \sqrt{6}) 2 \sqrt{3}}{2 \sqrt{3} \times 2 \sqrt{3}}=\mathrm{k}
\frac{18 \sqrt{3}-12 \sqrt{2}}{12}=\mathrm{k}
3√3 – 2√2 = k
\frac{18-2 \sqrt{3} \mathrm{k}-4 \sqrt{6}}{4}=0
18 – 2√3 k – 4√6 = 0
18 – 4√6 = 2√3k
\frac{18-4 \sqrt{6}}{2 \sqrt{3}}=\mathrm{k}
\frac{(18-4 \sqrt{6}) 2 \sqrt{3}}{2 \sqrt{3} \times 2 \sqrt{3}}=\mathrm{k}
\frac{18 \sqrt{3}-12 \sqrt{2}}{12}=\mathrm{k}
3√3 – 2√2 = k
(iv) ax2 + bx – 10 =
0
Put x=-\frac{\sqrt{3}}{2}
a\left(-\frac{2}{5}\right)^{2}+b\left(-\frac{2}{5}\right)-10=0
\frac{4 a}{25}-\frac{2 b}{5}-10=0
\frac{4 a-10 b-250}{25}=0
4a – 10b – 250 = 0
4a – 10b = 250 (1)
Put x = 3/5
ax2 + bx – 10 = 0
a\left(\frac{5}{3}\right)^{2}+b\left(\frac{5}{3}\right)-10=0
\frac{25 a}{9}+\frac{5 b}{3}-10=0
\frac{25 a+15 b-90}{9}=0
25a + 15b – 90 = 0 (divide by 5)
5a + 3b = 18 (2)
Eliminate (1) and (2)
4a – 10b = 250 ×5
5a + 3b = 18 ×4
Put x=-\frac{\sqrt{3}}{2}
a\left(-\frac{2}{5}\right)^{2}+b\left(-\frac{2}{5}\right)-10=0
\frac{4 a}{25}-\frac{2 b}{5}-10=0
\frac{4 a-10 b-250}{25}=0
4a – 10b – 250 = 0
4a – 10b = 250 (1)
Put x = 3/5
ax2 + bx – 10 = 0
a\left(\frac{5}{3}\right)^{2}+b\left(\frac{5}{3}\right)-10=0
\frac{25 a}{9}+\frac{5 b}{3}-10=0
\frac{25 a+15 b-90}{9}=0
25a + 15b – 90 = 0 (divide by 5)
5a + 3b = 18 (2)
Eliminate (1) and (2)
4a – 10b = 250 ×5
5a + 3b = 18 ×4
\begin{aligned}20a-50b&=1250\\-2a-12b&=-72\\ \hline
-62b&=1178\end{aligned}
b = –19
Put b = –19 in (1)
4a – 10b = 250
4a – 10(–19) = 250
4a + 190 = 250
4a = 60
a = 15
Sol :
2x– 2x – 3x + 3 = 0
2x (x– 1) – 3(x – 1) = 0
(2x – 3) (x – 1) = 0
2x – 3 = 0
x=\frac{3}{2}
x – 1 = 0
x =1
Therefore, the roots of the equation are \frac{3}{2}, 1.
Sol :
3x2 – √6 x – √6 x + 2 = 0
3x2 – √2√3x – √2√3x + 2 = 0
√3x(√3x – √2) – √2 (√3x – √2) = 0
√3x – √2 = 0 √3x – √2 = 0
\mathrm{x}=\frac{\sqrt{2}}{\sqrt{3}}, \mathrm{x}=\frac{\sqrt{2}}{\sqrt{3}}
Therefore, the roots of the equation are \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}
Sol :
3x2 — 15x + x — 5 = 0
3x (x – 5) + (x – 5) = 0
(3x + 1) (x – 5) = 0
3x + 1 = 0 x – 5 = 0
x=-\frac{1}{3}, x=5
Therefore, the roots of the equation are -\frac{1}{3}, 5
Sol :
Now, to find the roots by factorisation, we need to factorise 10 such that the sum is 10 and the product is
Put b = –19 in (1)
4a – 10b = 250
4a – 10(–19) = 250
4a + 190 = 250
4a = 60
a = 15
Question 14 A
Find the roots of the following quadratic equations by factorisation:
2x2 — 5x + 3 = 0Sol :
2x– 2x – 3x + 3 = 0
2x (x– 1) – 3(x – 1) = 0
(2x – 3) (x – 1) = 0
2x – 3 = 0
x=\frac{3}{2}
x – 1 = 0
x =1
Therefore, the roots of the equation are \frac{3}{2}, 1.
Question 14 B
Find the roots of the following quadratic equations by factorisation:
3x2 —2√6x +2 = 0Sol :
3x2 – √6 x – √6 x + 2 = 0
3x2 – √2√3x – √2√3x + 2 = 0
√3x(√3x – √2) – √2 (√3x – √2) = 0
√3x – √2 = 0 √3x – √2 = 0
\mathrm{x}=\frac{\sqrt{2}}{\sqrt{3}}, \mathrm{x}=\frac{\sqrt{2}}{\sqrt{3}}
Therefore, the roots of the equation are \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}
Question 14 C
Find the roots of the following quadratic equations by factorisation:
3x2 — 14x — 5 = 0Sol :
3x2 — 15x + x — 5 = 0
3x (x – 5) + (x – 5) = 0
(3x + 1) (x – 5) = 0
3x + 1 = 0 x – 5 = 0
x=-\frac{1}{3}, x=5
Therefore, the roots of the equation are -\frac{1}{3}, 5
Question 14 D
Find the roots of the following quadratic equations by factorisation:
\sqrt{3} \mathrm{x}^{2}+10 \mathrm{x}+7 \sqrt{3}=0Sol :
Now, to find the roots by factorisation, we need to factorise 10 such that the sum is 10 and the product is
7√3×√3=21
We can do that by 7 and 3.
So,
√3x2 + 10x + 7√3 = 0
√3x2 + 3x + 7x + 7√3 = 0
√3x(x + √3) + 7(x + √3) = 0
(√3x + 7)(x + √3) = 0
(√3x + 7) = 0
\sqrt{3} x=-7
x=-\frac{7}{\sqrt{3}}
Or
x + √3 = 0
x = -√3
Hence, the solutions of the given quadratic equations are -√3 and -\frac{7}{\sqrt{3}}.
Sol :
√7 y2 – 13y + 7y — 13 √7 = 0
y (√7 y – 13) + √7 (√7 y – 13) = 0
(√7 y – 13) (y + √7) = 0
√7 y – 13 = 0 y + √7 = 0
\mathrm{y}=\frac{13}{\sqrt{7}} y = –√7
Rationalise
\mathrm{y}=\frac{13}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}
\mathrm{y}=\frac{13 \sqrt{7}}{7}
Therefore, the roots of the equation are \frac{13 \sqrt{7}}{7}-\sqrt{7}
Sol :
4x2 – {2(a2 + b2) + 2(a2 – b2)}x + (a2 + b2) (a2 – b2) = 0
4x2 – 2(a2 + b2)x + 2(a2 – b2) x + (a2 + b2) (a2 – b2) = 0
2x {2x – (a2 + b2)} – (a2 – b2) {2x – (a2 + b2)} = 0
{2x – (a2 – b2)} {2x – (a2 + b2)} = 0
2x – (a2 – b2) = 0 2x – (a2 + b2) = 0
2x = (a2 – b2) 2x = (a2 + b2)
x=\frac{a^{2}-b^{2}}{2}, x=\frac{a^{2}+b^{2}}{2}
Therefore, the roots of the equation are \frac{a^{2}-b^{2}}{2} ,\frac{a^{2}+b^{2}}{2}
b2x {a2x + 1} – 1 {a2x + 1} = 0
(b2x – 1) (a2x + 1) = 0
b2x – 1 = 0 a2x + 1 = 0
x=\frac{1}{b^{2}}, x=\frac{-1}{a^{2}}
Therefore, the roots of the equation are \frac{1}{b^{2}}, \frac{-1}{a^{2}}
(6x)2 – 2 (6x)a + a2 — b2 = 0
Using Identity:
(x– y)2 = x2 + y2 – 2xy
Here, (6x–a)2 = (6x)2 – 2 (6x) a + a2
(6x – a)2 – b2 = 0
Using identity:
x2 – y2 = (a+ b) (a – b)
(6x –a + b) (6x – a – b) =0
6x = a – b 6x = a + b
x=\frac{a-b}{6},x=\frac{a+b}{6}
Therefore, the roots of the equation are\frac{a-b}{6}, \frac{a+b}{6}
2x (5ax – 3) + 3 (5ax – 3) = 0
2x+3 =0 5ax – 3 = 0
x=-\frac{3}{2}, x=\frac{3}{50}
Therefore, roots of the equation are\frac{3}{50},-\frac{3}{2}
12abx2 — 9a2 x— 8b2x — 6ab = 0
3ax (4bx – 3a) + 2b (4bx – 3a) = 0
3ax + 2b =0 4bx – 3a =0
x=-\frac{2 b}{3 a}, x=\frac{3 a}{4 b}
Therefore, roots of the equation are -\frac{2 \mathrm{b}}{3 \mathrm{a}}, \frac{3 \mathrm{a}}{4 \mathrm{b}}
4x2 — 2a2x+ 2b2x + a2b2 = 0
2x (2x – a2) – b2 (2x – a2) = 0
2x – b2 = 0 2x – a2 = 0
x=\frac{b^{2}}{2} ,x=\frac{a^{2}}{2}
Sol :
5x2 – 6x – 2 = 0
Dividing by 5
x^{2}-\frac{6 x}{5}-\frac{2}{5}=0
We know
(a – b)2 = a2 – 2ab + b2
We can do that by 7 and 3.
So,
√3x2 + 10x + 7√3 = 0
√3x2 + 3x + 7x + 7√3 = 0
√3x(x + √3) + 7(x + √3) = 0
(√3x + 7)(x + √3) = 0
(√3x + 7) = 0
\sqrt{3} x=-7
x=-\frac{7}{\sqrt{3}}
Or
x + √3 = 0
x = -√3
Hence, the solutions of the given quadratic equations are -√3 and -\frac{7}{\sqrt{3}}.
Question 14 E
Find the roots of the following quadratic equations by factorisation:
√7 y2 – 6y — 13√7 = 0Sol :
√7 y2 – 13y + 7y — 13 √7 = 0
y (√7 y – 13) + √7 (√7 y – 13) = 0
(√7 y – 13) (y + √7) = 0
√7 y – 13 = 0 y + √7 = 0
\mathrm{y}=\frac{13}{\sqrt{7}} y = –√7
Rationalise
\mathrm{y}=\frac{13}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}
\mathrm{y}=\frac{13 \sqrt{7}}{7}
Therefore, the roots of the equation are \frac{13 \sqrt{7}}{7}-\sqrt{7}
Question 14 F
Find the roots of the following quadratic equations by factorisation:
4x2 — 4a2x + a4 — b4 =
0Sol :
4x2 – {2(a2 + b2) + 2(a2 – b2)}x + (a2 + b2) (a2 – b2) = 0
4x2 – 2(a2 + b2)x + 2(a2 – b2) x + (a2 + b2) (a2 – b2) = 0
2x {2x – (a2 + b2)} – (a2 – b2) {2x – (a2 + b2)} = 0
{2x – (a2 – b2)} {2x – (a2 + b2)} = 0
2x – (a2 – b2) = 0 2x – (a2 + b2) = 0
2x = (a2 – b2) 2x = (a2 + b2)
x=\frac{a^{2}-b^{2}}{2}, x=\frac{a^{2}+b^{2}}{2}
Therefore, the roots of the equation are \frac{a^{2}-b^{2}}{2} ,\frac{a^{2}+b^{2}}{2}
Question 15 A
a2b2x2 + b2x – a2x — 1 = 0, a ≠ 0, b ≠
0
Sol :b2x {a2x + 1} – 1 {a2x + 1} = 0
(b2x – 1) (a2x + 1) = 0
b2x – 1 = 0 a2x + 1 = 0
x=\frac{1}{b^{2}}, x=\frac{-1}{a^{2}}
Therefore, the roots of the equation are \frac{1}{b^{2}}, \frac{-1}{a^{2}}
Question 15 B
36x2 — 12ax + (a2 — b2) = 0
Sol :(6x)2 – 2 (6x)a + a2 — b2 = 0
Using Identity:
(x– y)2 = x2 + y2 – 2xy
Here, (6x–a)2 = (6x)2 – 2 (6x) a + a2
(6x – a)2 – b2 = 0
Using identity:
x2 – y2 = (a+ b) (a – b)
(6x –a + b) (6x – a – b) =0
6x = a – b 6x = a + b
x=\frac{a-b}{6},x=\frac{a+b}{6}
Therefore, the roots of the equation are\frac{a-b}{6}, \frac{a+b}{6}
Question 15 C
10ax2 — 6x + 15ax – 9 = 0, a ≠ 0
Sol :2x (5ax – 3) + 3 (5ax – 3) = 0
2x+3 =0 5ax – 3 = 0
x=-\frac{3}{2}, x=\frac{3}{50}
Therefore, roots of the equation are\frac{3}{50},-\frac{3}{2}
Question 15 D
12abx2 — (9a2 — 8b2) x — 6ab = 0
Sol :12abx2 — 9a2 x— 8b2x — 6ab = 0
3ax (4bx – 3a) + 2b (4bx – 3a) = 0
3ax + 2b =0 4bx – 3a =0
x=-\frac{2 b}{3 a}, x=\frac{3 a}{4 b}
Therefore, roots of the equation are -\frac{2 \mathrm{b}}{3 \mathrm{a}}, \frac{3 \mathrm{a}}{4 \mathrm{b}}
Question 15 E
4x2 — 2 (a2 + b2) x + a2b2 =
0
Sol :4x2 — 2a2x+ 2b2x + a2b2 = 0
2x (2x – a2) – b2 (2x – a2) = 0
2x – b2 = 0 2x – a2 = 0
x=\frac{b^{2}}{2} ,x=\frac{a^{2}}{2}
Question 16 A
Find the roots of the following quadratic equations, if they exist by the method of completing the
square:
5x2 – 6x – 2 = 0Sol :
5x2 – 6x – 2 = 0
Dividing by 5
x^{2}-\frac{6 x}{5}-\frac{2}{5}=0
We know
(a – b)2 = a2 – 2ab + b2
Here, a=x and –2ab =-\frac{6
x}{5}
–2xb = -\frac{6 x}{5}(∵ a =x)
–2b = -\frac{6}{5}
\mathrm{b}=-\frac{6}{5 \times(-2)}
\mathrm{b}=\frac{3}{5}
∴ Equation becomes
–2xb = -\frac{6 x}{5}(∵ a =x)
–2b = -\frac{6}{5}
\mathrm{b}=-\frac{6}{5 \times(-2)}
\mathrm{b}=\frac{3}{5}
∴ Equation becomes
x^{2}-\frac{6 x}{5}-\frac{2}{5}=0
Add and subtract \left(\frac{3}{5}\right)^{2}
x^{2}-\frac{6 x}{5}-\frac{2}{5}+\left(\frac{3}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=0
x^{2}-\frac{6 x}{5}+\left(\frac{3}{5}\right)^{2}-\frac{2}{5}-\left(\frac{3}{5}\right)^{2}=0
\left(x-\frac{3}{5}\right)^{2}=\left(\frac{3}{5}\right)^{2}+\frac{2}{5}
\left(x-\frac{3}{5}\right)^{2}=\frac{9}{25}+\frac{2}{5}
\left(x-\frac{3}{5}\right)^{2}=\frac{9+2(5)}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{9+10}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{19}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{(\sqrt{19})^{2}}{(5)^{2}}
Add and subtract \left(\frac{3}{5}\right)^{2}
x^{2}-\frac{6 x}{5}-\frac{2}{5}+\left(\frac{3}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=0
x^{2}-\frac{6 x}{5}+\left(\frac{3}{5}\right)^{2}-\frac{2}{5}-\left(\frac{3}{5}\right)^{2}=0
\left(x-\frac{3}{5}\right)^{2}=\left(\frac{3}{5}\right)^{2}+\frac{2}{5}
\left(x-\frac{3}{5}\right)^{2}=\frac{9}{25}+\frac{2}{5}
\left(x-\frac{3}{5}\right)^{2}=\frac{9+2(5)}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{9+10}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{19}{25}
\left(x-\frac{3}{5}\right)^{2}=\frac{(\sqrt{19})^{2}}{(5)^{2}}
Canceling squares
both sides
\left(x-\frac{3}{5}\right)=\pm \frac{\sqrt{19}}{5}
\left(x-\frac{3}{5}\right)=\pm \frac{\sqrt{19}}{5}
Solving
\left(x-\frac{3}{5}\right)=\frac{\sqrt{19}}{5}\left(x-\frac{3}{5}\right)=-\frac{\sqrt{19}}{5}
x=\frac{\sqrt{19}}{5}+\frac{3}{5} x=\frac{-\sqrt{19}}{5}+\frac{3}{5}
x=\frac{\sqrt{19}+3}{5} ,x=\frac{-\sqrt{19}+3}{5}
So, \mathrm{x}=\frac{\sqrt{19}+3}{5} and x=\frac{-\sqrt{19}+3}{5} are the roots of the equation.
Sol :
2x2 – 5x + 3 = 0
Dividing by 2
x^{2}-\frac{5 x}{2}+\frac{3}{2}=0
x^{2}-\frac{5 x}{2}=-\frac{3}{2}
\left(x-\frac{3}{5}\right)=\frac{\sqrt{19}}{5}\left(x-\frac{3}{5}\right)=-\frac{\sqrt{19}}{5}
x=\frac{\sqrt{19}}{5}+\frac{3}{5} x=\frac{-\sqrt{19}}{5}+\frac{3}{5}
x=\frac{\sqrt{19}+3}{5} ,x=\frac{-\sqrt{19}+3}{5}
So, \mathrm{x}=\frac{\sqrt{19}+3}{5} and x=\frac{-\sqrt{19}+3}{5} are the roots of the equation.
Question 16 B
Find the roots of the following quadratic equations, if they exist by the method of completing the
square:
2x2 – 5x + 3 = 0Sol :
2x2 – 5x + 3 = 0
Dividing by 2
x^{2}-\frac{5 x}{2}+\frac{3}{2}=0
x^{2}-\frac{5 x}{2}=-\frac{3}{2}
Add a coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-\frac{5 x}{2}+\left(\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{5}{4}\right)^{2}
\left(x-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\frac{25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{-24+25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}
x-\frac{5}{4}=\sqrt{\frac{1}{16}}
x=\frac{1}{4}+\frac{5}{4}, x=-\frac{1}{4}+\frac{5}{4}
x=\frac{6}{4}=\frac{3}{2}, x=\frac{4}{4}=1
Sol :
Dividing by 9
x^{2}-\frac{15 x}{9}+\frac{6}{9}=0
x^{2}-\frac{15 x}{9}=-\frac{2}{3}
x^{2}-\frac{5 x}{2}+\left(\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{5}{4}\right)^{2}
\left(x-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\frac{25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{-24+25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}
x-\frac{5}{4}=\sqrt{\frac{1}{16}}
x=\frac{1}{4}+\frac{5}{4}, x=-\frac{1}{4}+\frac{5}{4}
x=\frac{6}{4}=\frac{3}{2}, x=\frac{4}{4}=1
Question 16 C
Find the roots of the following quadratic equations, if they exist by the method of completing the
square:
9x2 – 15x + 6 = 0Sol :
Dividing by 9
x^{2}-\frac{15 x}{9}+\frac{6}{9}=0
x^{2}-\frac{15 x}{9}=-\frac{2}{3}
Add a coefficient
of \left(\frac{x}{2}\right)^{2} to both
sides
\mathrm{x}^{2}-\frac{15 \mathrm{x}}{9}+\left(\frac{15}{18}\right)^{2}=-\frac{2}{3}+\left(\frac{15}{18}\right)^{2}
\left(x-\frac{15}{18}\right)^{2}=-\frac{2}{3}+\frac{25}{36}
\left(x-\frac{15}{18}\right)^{2}=\frac{25-24}{36}
\left(x-\frac{15}{18}\right)^{2}=\frac{1}{36}
x-\frac{15}{18}=\frac{\sqrt{1}}{\sqrt{36}}
x=\frac{1}{6}+\frac{5}{6}, x=-\frac{1}{6}+\frac{5}{6}
\mathrm{x}^{2}-\frac{15 \mathrm{x}}{9}+\left(\frac{15}{18}\right)^{2}=-\frac{2}{3}+\left(\frac{15}{18}\right)^{2}
\left(x-\frac{15}{18}\right)^{2}=-\frac{2}{3}+\frac{25}{36}
\left(x-\frac{15}{18}\right)^{2}=\frac{25-24}{36}
\left(x-\frac{15}{18}\right)^{2}=\frac{1}{36}
x-\frac{15}{18}=\frac{\sqrt{1}}{\sqrt{36}}
x=\frac{1}{6}+\frac{5}{6}, x=-\frac{1}{6}+\frac{5}{6}
x=\frac{6}{6}=1, x=\frac{4}{6}=\frac{2}{3}
Sol :
x2 – 9x = –18
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-9 x+\left(\frac{9}{2}\right)^{2}=-18+\left(\frac{9}{2}\right)^{2}
\left(x-\frac{9}{2}\right)^{2}=-18+\frac{81}{4}
\left(x-\frac{9}{2}\right)^{2}=\frac{-72+81}{4}
\left(x-\frac{9}{2}\right)^{2}=\frac{9}{4}
x-\frac{9}{2}=\frac{\sqrt{9}}{\sqrt{4}}
x-\frac{9}{2}=\frac{\pm 3}{2}
x=-\frac{3}{2}+\frac{9}{2}, x=\frac{3}{2}+\frac{9}{2}
x=\frac{6}{2}=3, x=\frac{12}{2}=6
Sol :
x^{2}+\frac{x}{2}+2=0
x^{2}+\frac{x}{2}=-2
Question 16 D
Find the roots of the following quadratic equations, if they exist by the method of completing the
square:
x2 – 9x + 18 = 0Sol :
x2 – 9x = –18
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-9 x+\left(\frac{9}{2}\right)^{2}=-18+\left(\frac{9}{2}\right)^{2}
\left(x-\frac{9}{2}\right)^{2}=-18+\frac{81}{4}
\left(x-\frac{9}{2}\right)^{2}=\frac{-72+81}{4}
\left(x-\frac{9}{2}\right)^{2}=\frac{9}{4}
x-\frac{9}{2}=\frac{\sqrt{9}}{\sqrt{4}}
x-\frac{9}{2}=\frac{\pm 3}{2}
x=-\frac{3}{2}+\frac{9}{2}, x=\frac{3}{2}+\frac{9}{2}
x=\frac{6}{2}=3, x=\frac{12}{2}=6
Question 16 E
Find the roots of the following quadratic equations, if they exist by the method of completing the
square:
2x2 + x + 4 = 0Sol :
x^{2}+\frac{x}{2}+2=0
x^{2}+\frac{x}{2}=-2
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both
sides
x^{2}+\frac{x}{2}+\left(\frac{1}{4}\right)^{2}=-2+\left(\frac{1}{4}\right)^{2}
\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}
\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}
\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}
x+\frac{1}{4}=\sqrt{-\frac{31}{16}}
Since root cannot be negative
x^{2}+\frac{x}{2}+\left(\frac{1}{4}\right)^{2}=-2+\left(\frac{1}{4}\right)^{2}
\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}
\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}
\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}
x+\frac{1}{4}=\sqrt{-\frac{31}{16}}
Since root cannot be negative
Therefore, no real roots exist.
Sol :
x^{2}-\frac{5 x}{2}+\frac{3}{2}=0
x^{2}-\frac{5 x}{2}=-\frac{3}{2}
Question 17 A
Find the roots of each of the following quadratic equations if they exist by the method of completing the
squares:
2x2 – 5x + 3 = 0Sol :
x^{2}-\frac{5 x}{2}+\frac{3}{2}=0
x^{2}-\frac{5 x}{2}=-\frac{3}{2}
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-\frac{5 x}{2}+\left(\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{5}{4}\right)^{2}
\left(x-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\frac{25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{-24+25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}
x-\frac{5}{4}=\frac{\sqrt{1}}{\sqrt{16}}
x=\frac{1+5}{4}, x=\frac{-1+5}{4}
x=\frac{6}{4}=\frac{3}{2}, x=\frac{4}{4}=1
Sol :
x2 – 6x = –4
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x2 – 6x + (3)2 = –4 + (3)2
(x – 3)2 = –4 + 9
x – 3 = √5
x = ±√5 + 3
Sol :
Divide by √5
x^{2}+\frac{9 x}{\sqrt{5}}=-4
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+\frac{9 x}{\sqrt{5}}+\left(\frac{9}{2 \sqrt{5}}\right)^{2}=-4+\left(\frac{9}{2 \sqrt{5}}\right)^{2}
\left(x+\frac{9}{2 \sqrt{5}}\right)^{2}=-4+\frac{81}{20}
\left(x+\frac{9}{2 \sqrt{5}}\right)^{2}=\frac{-80+81}{20}
x+\frac{9}{2 \sqrt{5}}=\frac{\sqrt{1}}{\sqrt{20}}
\mathrm{x}=\frac{1}{2 \sqrt{5}}-\frac{9}{2 \sqrt{5}}, \mathrm{x}=\frac{-1}{2 \sqrt{5}}-\frac{9}{2 \sqrt{5}}
\mathrm{x}=-\frac{8}{2 \sqrt{5}}=-\frac{4}{\sqrt{5}}, \mathrm{x}=-\frac{10}{2 \sqrt{5}}=-\frac{5}{\sqrt{5}}
x=-\frac{5}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=-\frac{5 \sqrt{5}}{5}=-\sqrt{5}
Sol :
Divide by 2
\mathrm{x}^{2}+\frac{\sqrt{15 \mathrm{x}}}{2}=-\frac{\sqrt{2}}{2}
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
\mathrm{x}^{2}+\frac{\sqrt{15 \mathrm{x}}}{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}=-\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}
\left(x+\frac{\sqrt{15}}{4}\right)^{2}=-\frac{\sqrt{2}}{2}+\frac{15}{16}
\left(x+\frac{\sqrt{15}}{4}\right)^{2}=\frac{-8 \sqrt{2}+15}{16}
Since root cannot be negative
Therefore, it has no real roots.
Sol :
x2 + x = –3
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-3+\left(\frac{1}{2}\right)^{2}
\left(x+\frac{1}{2}\right)^{2}=-3+\frac{1}{4}
\left(x+\frac{1}{2}\right)^{2}=\frac{-12+1}{4}
\left(x+\frac{1}{2}\right)^{2}=\frac{-11}{4}
x+\frac{1}{2}=\sqrt{-\frac{11}{4}}
Since root cannot be negative
Therefore, it has no real roots.
Sol :
Dividing by 5
x^{2}-\frac{24 x}{5}-\frac{5}{5}=0
x^{2}-\frac{24 x}{5}=1
x^{2}-\frac{5 x}{2}+\left(\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{5}{4}\right)^{2}
\left(x-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\frac{25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{-24+25}{16}
\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}
x-\frac{5}{4}=\frac{\sqrt{1}}{\sqrt{16}}
x=\frac{1+5}{4}, x=\frac{-1+5}{4}
x=\frac{6}{4}=\frac{3}{2}, x=\frac{4}{4}=1
Question 17 B
Find the roots of each of the following quadratic equations if they exist by the method of completing the
squares:
x2 – 6x + 4 = 0Sol :
x2 – 6x = –4
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x2 – 6x + (3)2 = –4 + (3)2
(x – 3)2 = –4 + 9
x – 3 = √5
x = ±√5 + 3
Question 17 C
Find the roots of each of the following quadratic equations if they exist by the method of completing the
squares:
√5x2 + 9x + 4√5 = 0Sol :
Divide by √5
x^{2}+\frac{9 x}{\sqrt{5}}=-4
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+\frac{9 x}{\sqrt{5}}+\left(\frac{9}{2 \sqrt{5}}\right)^{2}=-4+\left(\frac{9}{2 \sqrt{5}}\right)^{2}
\left(x+\frac{9}{2 \sqrt{5}}\right)^{2}=-4+\frac{81}{20}
\left(x+\frac{9}{2 \sqrt{5}}\right)^{2}=\frac{-80+81}{20}
x+\frac{9}{2 \sqrt{5}}=\frac{\sqrt{1}}{\sqrt{20}}
\mathrm{x}=\frac{1}{2 \sqrt{5}}-\frac{9}{2 \sqrt{5}}, \mathrm{x}=\frac{-1}{2 \sqrt{5}}-\frac{9}{2 \sqrt{5}}
\mathrm{x}=-\frac{8}{2 \sqrt{5}}=-\frac{4}{\sqrt{5}}, \mathrm{x}=-\frac{10}{2 \sqrt{5}}=-\frac{5}{\sqrt{5}}
x=-\frac{5}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=-\frac{5 \sqrt{5}}{5}=-\sqrt{5}
Question 17 D
Find the roots of each of the following quadratic equations if they exist by the method of completing the
squares:
2x2 + √15 x + √2 = 0Sol :
Divide by 2
\mathrm{x}^{2}+\frac{\sqrt{15 \mathrm{x}}}{2}=-\frac{\sqrt{2}}{2}
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
\mathrm{x}^{2}+\frac{\sqrt{15 \mathrm{x}}}{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}=-\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}
\left(x+\frac{\sqrt{15}}{4}\right)^{2}=-\frac{\sqrt{2}}{2}+\frac{15}{16}
\left(x+\frac{\sqrt{15}}{4}\right)^{2}=\frac{-8 \sqrt{2}+15}{16}
Since root cannot be negative
Therefore, it has no real roots.
Question 17 E
Find the roots of each of the following quadratic equations if they exist by the method of completing the
squares:
x2 + x + 3 = 0Sol :
x2 + x = –3
Add the coefficient of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-3+\left(\frac{1}{2}\right)^{2}
\left(x+\frac{1}{2}\right)^{2}=-3+\frac{1}{4}
\left(x+\frac{1}{2}\right)^{2}=\frac{-12+1}{4}
\left(x+\frac{1}{2}\right)^{2}=\frac{-11}{4}
x+\frac{1}{2}=\sqrt{-\frac{11}{4}}
Since root cannot be negative
Therefore, it has no real roots.
Question 18 A
Solve the following equations by the method of completion of a square.
5x2 – 24x – 5 = 0Sol :
Dividing by 5
x^{2}-\frac{24 x}{5}-\frac{5}{5}=0
x^{2}-\frac{24 x}{5}=1
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-\frac{24
x}{5}+\left(\frac{24}{10}\right)^{2}=1+\left(\frac{24}{10}\right)^{2}
\left(x-\frac{24}{10}\right)^{2}=1+5.76
x-\frac{24}{10}=\sqrt{6.76}
x = 2.6 + 2.4 x = –2.6 + 2.4
x = 5 x = – 0.2
Sol :
Divide by 7
x^{2}-\frac{13 x}{7}=\frac{2}{7}
\left(x-\frac{24}{10}\right)^{2}=1+5.76
x-\frac{24}{10}=\sqrt{6.76}
x = 2.6 + 2.4 x = –2.6 + 2.4
x = 5 x = – 0.2
Question 18 B
Solve the following equations by the method of completion of a square.
7x2 – 13x – 2 = 0Sol :
Divide by 7
x^{2}-\frac{13 x}{7}=\frac{2}{7}
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}-\frac{13 x}{7}+\left(\frac{13}{14}\right)^{2}=\frac{2}{7}+\left(\frac{13}{14}\right)^{2}
\left(x-\frac{13}{14}\right)^{2}=\frac{2}{7}+\frac{169}{196}
\left(x-\frac{13}{14}\right)^{2}=\frac{56+169}{196}
\left(x-\frac{13}{14}\right)^{2}=\frac{225}{196}
x-\frac{13}{14}=\frac{\sqrt{225}}{\sqrt{196}}
x-\frac{13}{14}=\frac{15}{14}
x=\frac{15+13}{14} ,x=\frac{-15+13}{14}
x=\frac{28}{14}=2 x=-\frac{2}{14}=-\frac{1}{7}
Sol :
Divide by 15
x^{2}+\frac{53 x}{15}=-\frac{42}{15}
x^{2}-\frac{13 x}{7}+\left(\frac{13}{14}\right)^{2}=\frac{2}{7}+\left(\frac{13}{14}\right)^{2}
\left(x-\frac{13}{14}\right)^{2}=\frac{2}{7}+\frac{169}{196}
\left(x-\frac{13}{14}\right)^{2}=\frac{56+169}{196}
\left(x-\frac{13}{14}\right)^{2}=\frac{225}{196}
x-\frac{13}{14}=\frac{\sqrt{225}}{\sqrt{196}}
x-\frac{13}{14}=\frac{15}{14}
x=\frac{15+13}{14} ,x=\frac{-15+13}{14}
x=\frac{28}{14}=2 x=-\frac{2}{14}=-\frac{1}{7}
Question 18 C
Solve the following equations by the method of completion of a square.
15x2 + 53x + 42 = 0Sol :
Divide by 15
x^{2}+\frac{53 x}{15}=-\frac{42}{15}
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+\frac{53 x}{15}+\left(\frac{53}{30}\right)^{2}=-\frac{42}{15}+\left(\frac{53}{30}\right)^{2}
\left(x+\frac{53}{30}\right)^{2}=-\frac{42}{15}+\frac{2809}{900}
\left(x+\frac{53}{30}\right)^{2}=\frac{-2520+2809}{900}
x+\frac{53}{30}=\frac{\sqrt{289}}{\sqrt{900}}
x=\frac{17}{30}-\frac{53}{30}, x=-\frac{17}{30}-\frac{53}{30}
x=-\frac{6}{5} ,x=-\frac{7}{3}
Sol :
Divide by 7
x^{2}+\frac{2 x}{7}=\frac{5}{7}
x^{2}+\frac{53 x}{15}+\left(\frac{53}{30}\right)^{2}=-\frac{42}{15}+\left(\frac{53}{30}\right)^{2}
\left(x+\frac{53}{30}\right)^{2}=-\frac{42}{15}+\frac{2809}{900}
\left(x+\frac{53}{30}\right)^{2}=\frac{-2520+2809}{900}
x+\frac{53}{30}=\frac{\sqrt{289}}{\sqrt{900}}
x=\frac{17}{30}-\frac{53}{30}, x=-\frac{17}{30}-\frac{53}{30}
x=-\frac{6}{5} ,x=-\frac{7}{3}
Question 18 D
Solve the following equations by the method of completion of a square.
7x2 + 2x – 5 = 0Sol :
Divide by 7
x^{2}+\frac{2 x}{7}=\frac{5}{7}
Add the coefficient
of \left(\frac{x}{2}\right)^{2} to both sides
x^{2}+\frac{2 x}{7}+\left(\frac{2}{14}\right)^{2}=\frac{5}{7}+\left(\frac{2}{14}\right)^{2}
\left(x+\frac{2}{14}\right)^{2}=\frac{5}{7}+\frac{4}{196}
\left(x+\frac{2}{14}\right)^{2}=\frac{140+4}{196}
\left(x+\frac{2}{14}\right)^{2}=\frac{144}{196}
x+\frac{2}{14}=\frac{\sqrt{144}}{\sqrt{196}}
x+\frac{2}{14}=\frac{12}{14}
x=\frac{12-2}{14} ,x=\frac{-12-2}{14}
x=\frac{10}{14}=\frac{5}{7} ,x=-\frac{14}{14}=-1
x^{2}+\frac{2 x}{7}+\left(\frac{2}{14}\right)^{2}=\frac{5}{7}+\left(\frac{2}{14}\right)^{2}
\left(x+\frac{2}{14}\right)^{2}=\frac{5}{7}+\frac{4}{196}
\left(x+\frac{2}{14}\right)^{2}=\frac{140+4}{196}
\left(x+\frac{2}{14}\right)^{2}=\frac{144}{196}
x+\frac{2}{14}=\frac{\sqrt{144}}{\sqrt{196}}
x+\frac{2}{14}=\frac{12}{14}
x=\frac{12-2}{14} ,x=\frac{-12-2}{14}
x=\frac{10}{14}=\frac{5}{7} ,x=-\frac{14}{14}=-1
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