Exercise
7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
Exercise 7.4
Question 1 A
Write the discriminate of each of the following quadratic equation:
x2 + 4x + 3 =0Sol :
d = b2 – 4ac
d = (4)2 – 4 (1) (3)
d = 16 – 12
d = 4
Question 1 B
Write the discriminate of each of the following quadratic equation:
4x2 + 5x + 7 = 0Sol :
d = b2 – 4ac
d = (5)2 – 4 (4) (7)
d = 25 – 112
d = –87
Question 1 C
Write the discriminate of each of the following quadratic equation:
2x2 + 4x + 5 = 0Sol :
d = b2 – 4ac
d = (4)2 – 4 (2) (5)
d = 16 – 40
d = –24
Question 1 D
Write the discriminate of each of the following quadratic equation:
3x2 + 5x + 6 = 0Sol :
d = b2 – 4ac
d = (5)2 – 4 (3) (6)
d = 25 – 72
d = –47
Question 1 E
Write the discriminate of each of the following quadratic equation:
√3 x2 — 2√2 – 2√3 =
0Sol :
d = b2– 4ac
d = (–2√2)2 – 4 (√3) (–2√3)
d = 8 + 24
d = 32
Question 2 A
Examine whether the following quadratic equations have real roots or not:
7x2 + 8x — 1 = 0Sol :
TO CHECK REAL ROOTS, ‘d’ SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (8)22 – 4 (7) (–1)
d = 64 – 28
d = 92
Yes, roots are real.
Question 2 B
Examine whether the following quadratic equations have real roots or not:
2x2 + 3x + 4 = 0Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2– 4ac
d = (3)2 – 4(2)(4)
d = 9 – 32
d = – 23
No, roots aren’t real.
Question 2 C
Examine whether the following quadratic equations have real roots or not:
x2 — 12x — 16 = 0Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (–12)2 – 4 (1) (–16)
d = 144 – 64
d = 208
Yes, roots are real .
Question 2 D
Examine whether the following quadratic equations have real roots or not:
x2 + x – 1= 0Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (1)2 – 4 (1) (–1)
d = 1 + 4
d = 5
Yes, roots are real.
Question 2 E
Examine whether the following quadratic equations have real roots or not:
x2 — 10x + 2 = 0Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (–10)2 – 4 (1) (2)
d = 100 – 8
d = 92
Yes, roots are real.
Question 3 A
Find whether the following quadratic equations have a repeated root :
9x2— 12x + 4 = 0Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (–12)2 – 4(9)(4)
d = 144 – 144
d = 0
Yes, roots are repeated.
Question 3 B
Find whether the following quadratic equations have a repeated root :
y2 — 6y + 6 = 0Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (–6)2 – 4(1)(6)
d = 36 – 24
d = 12
∴ roots are not repeated.
Question 3 C
Find whether the following quadratic equations have a repeated root :
9x2 + 4x + 6 = 0Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (4)2 – 4 (9) (6)
d = 16 – 216
d = –200
∴ roots are not repeated.
Question 3 D
Find whether the following quadratic equations have a repeated root :
16y2— 40y + 25 = 0Sol :
REPEATED ROOTS MEAN d = 0 .
d = b2 – 4ac
d = (–40)2 – 4 (16) (25)
d = 1600 – 1600
d = 0
∴ roots are repeated.
Question 3 E
Find whether the following quadratic equations have a repeated root :
x2 + 6x + 9 = 0Sol :
REPEATED ROOTS MEAN d = 0 .
d = b2 – 4ac
d = (6)2 – 4(1)(9)
d = 36 – 36
d = 0
∴ roots are repeated.
d = b2 – 4ac
d = (6)2 – 4(1)(9)
d = 36 – 36
d = 0
∴ roots are repeated.
Question 4 A
Comment upon the nature of roots of the following equations:
4x2 + 7x + 2 = 0Sol :
d = b2 – 4ac
d = (7)2 – 4 (4) (2)
d = 49 – 32
d = 17
Since, d>0, roots are unique and real.
Question 4 B
Comment upon the nature of roots of the following equations:
x2 + 10x + 39 = 0Sol :
d = b2 – 4ac
d = (10)2 – 4 (1) (39)
d = 100 – 156
d = –56
Since, d < 0, no real roots exists.
Question 5 A
Without solving, determine whether the following equations have real roots or not. If yes, find
them:
2x2— 4x + 3 = 0Sol :
d = b2 – 4ac
d = (–4)2 – 4 (2) (3)
d = 16 – 24
d = –8
Since, d < 0, no real roots exist for the given equation.
Question 5 B
Without solving, determine whether the following equations have real roots or not. If yes, find
them:
$\mathrm{y}^{2}-\frac{2}{3} \mathrm{y}+\frac{1}{9}=0$d = b2 – 4ac
$\mathrm{d}=\left(\frac{-2}{3}\right)^{2}-4(1)\left(\frac{1}{9}\right)$
$d=\frac{4}{9}-\frac{4}{9}$
d = 0
Since, d = 0, roots are real and equal for the given equation.
$x=\frac{-b \pm \sqrt{d}}{2 a}$
$\mathrm{x}=\frac{-\left(-\frac{2}{3}\right) \pm \sqrt{0}}{2 \times 1}$
$x=\frac{2}{3} \times \frac{1}{2}$
$x=\frac{1}{3}$
Question 6 A
Without finding the roots, comment upon the nature of roots of each of the following quadratic
equations:
2x2— 6x + 3 = 0Sol :
d = b2 – 4ac
d = (–6)2 – 4 (2) (3)
d = 36 – 24
d = 12
∴ Roots are real and unique.
Question 6 B
Without finding the roots, comment upon the nature of roots of each of the following quadratic
equations:
2x2 — 5x — 3 = 0Sol :
d = b2 – 4ac
d = (–5)2 – 4 (2) (–3)
d = 25 + 24
d = 49
∴ Roots are real and unique.
Question 7 A
Find the value of k for which the quadratic equation
4x2 — 2 (k + 1) x + (k + 4) = 0 has equal roots.Sol :
Since roots are equal
∴ d=0 ….(1)
4x2 — 2 (k + 1) x + (k + 4) = 0
d = b2 – 4ac
d = (–2 (k + 1)2 – 4 (4) (k + 4)
d = (– 2k – 2)2 – 16k – 64
d = 4k2 + 4 + 8k – 16k – 64
(∵ (a – b)2 = a2 + b2 – 2ab)
d = 4k2 – 8k – 60
From (1), d = 0
∴ Equation will be:
4k2 – 8k – 60 = 0
Dividing by 4
k2 – 2k – 15 = 0
4k2 – 5k + 3k – 15 = 0
k (k – 5) + 3(k – 5) = 0
(k – 5) (k + 3) = 0
K – 5 = 0 k + 3 = 0
K = 5 k = –3
Question 7 B
Find the value of k, so that the quadratic equation
(k + 1) x2 – 2 (k — 1) x + 1 = 0 has equal roots.Sol :
Since roots are equal
∴ d=0 ….(1)
(k + 1)x2 — 2 (k – 1) x + 1 = 0
d = b2 – 4ac
d = (–2(k–1))2– 4(k+1)(1)
d = (–2k+2)2 – 4k – 4
d=4k2 + 4 – 8k – 4k – 4
(∵ (a + b)2 = a2 + b2 + 2ab)
d = 4k2 – 12k
From (1), d = 0
∴ Equation will be:
0 = 4k2 – 12k
4k2 = 12k
$\mathrm{k}^{2}=\frac{12}{4} \mathrm{k}$
k2 = 3k
k2 – 3k = 0
k(k – 3) = 0
k = 0 or k – 3 = 0
k = 3
∴ Values of k are 0, 3.
Question 8 A
For what values of k, does the following quadratic equation has equal roots.
9x2 + 8kx + 16 = 0Sol :
Since roots are equal
∴ d=0 (1)
9x2 + 8kx + 16 = 0
d = b2 – 4ac
d = (8k)2 – 4 (9) (16)
From (1), d = 0
∴ Equation will be:
0 = 64k2 – 576
$\mathrm{k}^{2}=\frac{576}{64}$
k2 = 9
k = ±√9
k = 3, –3
∴ values of k are –3, 3 .
Question 8 B
(k + 4)x2 + (k + 1)x + 1 = 0
Since roots are equal
∴ d=0 (1)
(k + 4)x2 + (k + 1)x + 1 = 0
d=b2–4ac
d = (k – 1)2– 4 (k + 4) (1)
d = (–2k + 2)2 – 4k – 4
d = k2 + 1+ 2k – 4k – 16
From (1), d = 0
∴ Equation will be:
0 = k2 + 1 + 2k – 4k – 16
k2 – 2k – 15 = 0
k2 – 5k + 3k – 15 = 0
k(k – 5) + 3 (k – 5) = 0
(k – 5) (k + 3) = 0
K – 5 = 0 k + 3 = 0
k = 5 k = –3
∴ Values of k are –3, 5.
Question 8 C
k2x2 — 2(2k — 1)x + 4 = 0
Sol :Since roots are equal
∴ d=0 (1)
k2x2 — 2(2k — 1)x + 4 = 0
d = b2 – 4ac
d = (–2(k–1))2 – 4 (k2) (4)
d = (–2k+2)2 – 4k – 4
d = (–4k+2)2 – 16k2
(∵ (a – b)2 = a2 + b2 – 2ab)
d = 16k2 – 16k + 4 – 16k2
d = –16k + 4
From (1), d = 0
∴ Equation will be:
0 = –16k + 4
16k = 4
∴ Values of k are is .
Question 9
If the roots of the equation (a — b)x2 + (b — c) x + (c — a) = 0 are equal, prove that 2a =
b + c.
Sol :Since roots are equal
∴ d=0 (1)
(a — b)x2 + (b — c) x + (c — a) = 0
d = b2 – 4ac
d = (b–c)2 – 4 (a–b) (c–a)
d = b2 + c2 – 2bc –4 [a (c – a) – b (c – a)]
d = b2 + c2 – 2bc – 4 [ac – a2 – bc + ba]
From (1), d = 0
∴ Equation will be:
0 = b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ba
b2 + c2 – (2a)2 2bc + 2c (–2a) + 2(–2a)b = 0
(b + c – 2a)2 = 0
(b + c – 2a) = 0
b + c = 2a
Hence proved.
Question 10
If — 5 is a root of the quadratic equation 2x2 + 2px — 15 = 0 and the quadratic equation p
(x2 + x) + k = 0 has equal roots, find the value of k.
Sol :2x2 + 2px — 15 = 0
Put x = –5
2(–5)2 + 2p(–5) — 15 = 0
50 – 10p – 15 = 0
35 = 10p
p = 3.5
Equation p (x2 + x) + k = 0 has equal roots i.e. d = 0
p (x2 + x) + k = 0
p = 3.5
3.5 (x2 + x) + k = 0
3.5x2 + 3.5x + k = 0
d = b2 – 4ac
d = (3.5)2 – 4(3.5)(k)
d = 12.25 – 14k
Putting d = 0
∴ Equation will be:
0 = 12.25 – 14k
14k = 12.25
$\mathrm{k}=\frac{12.25}{14}$
k = 0.875
Question 11 A
Find the values of k, for which the given equation has real roots:
2x2 — 10x + k = 0Roots are equal
∴ d = 0
d = b2 – 4ac
d = (–10)2 – 4 (2) (k)
d = 100 – 8k
Put d = 0
0 = 100 – 8k
8k = 100
$\mathrm{k}=\frac{100}{8}$
$\mathrm{k}=\frac{25}{2}$
Question 11 B
Find the values of k, for which the given equation has real roots:
kx2 — 6x — 2 = 0Sol :
Roots are equal
∴ d = 0
d = b2 – 4ac
d = (–6)2 – 4 (–2) (k)
d = 36 + 8k
Put d = 0
0 = 36 + 8k
–8k = 36
$\mathrm{k}=\frac{-36}{8}$
$\mathrm{k}=\frac{-9}{2}$
Question 11 C
Find the values of k, for which the given equation has real roots:
kx2+ 4x + 1 = 0Sol :
Roots are equal
∴ d = 0
d = b2 – 4ac
d = (4)2 – 4 (1) (k)
d = 16 – 4k
Put d = 0
0 = 16 – 4k
4k = 16
k = 4
Question 11 D
Find the values of k, for which the given equation has real roots:
kx2 – 2√5x + 4 = 0Sol :
Roots are equal
∴ d = 0
d = b2 – 4ac
d = (–2√5)2 – 4 (k) (4)
d = 20 – 16k
Put d = 0
0 = 20 – 16k
16k = 20
$\mathrm{k}=\frac{20}{16}$
$\mathrm{k}=\frac{5}{4}$
Question 11 E
Find the values of k, for which the given equation has real roots:
x2 + k(4x + k — 1) + 2 = 0Sol :
Roots are equal
∴ d = 0
d = b2 – 4ac
d = (4k)2 – 4 (k2 – k + 2) (1)
d = 16k2 – 4k2 + 4k – 8
Put d = 0
0 = 16k2 – 4k2 + 4k – 8
12k2 + 4k2 + 4k – 8 = 0 (divide by 4)
3k2 + k – 2 = 0
3k2 + 3k – 2k – 2 = 0
3k (k + 1) – 2k (k + 1) = 0
(3k–2k)(k+1) = 0
(3k–2k)=0 or (k+1) = 0
k = 0 k = –1
Question 12
Prove that the equation x2(a2 + b2) + 2x(ac + bd) +
(c2 + d2)= 0 has no real root, if ad ≠ bc.
Sol :x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0
d = b2 – 4ac
d = (2ac + 2bd)2 – 4 (a2 + b2) (c2 + d2)
d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2 (c2 + d2) + b2 (c2+d2)]
d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2c2 + a2d2 + b2c2 + b2d2]
d = 4a2c2 + 4b2d2 + 8abcd – 4a2c2 – 4a2d2 – 4b2c2 – 4b2 d2
d = 8abcd – 4a2d2 – 4b2c2
d = 8abcd – 4(a2d2 + b2 c2)
d = –4 (a2 d2 + b2c2 – 2abcd)
d = –4 [(ad + bc)2]
For ad ≠ bc
d= –4 × [value of (ad + bc)2]
∴ d is always negative
So, d < 0
The given equation has no real roots.
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