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Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0 is equal to their product.
5. Find k,
if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9
= 0 is equal to
their product.
Sol. 4x2 + 8kx + k + 9 = 0
Comparing with ax2 + bx + c =
0 we have a = 4, b = 8k, c = k + 9
Let α and β be the
roots of given quadratic equation.
α + β = α β ......(i) [Given]
(i.e.) –b/a =
c/a
∴ - 8k/4 = (k+9)/4
∴ -8k = k+9
∴ -8k – k = 9
∴ - 9k = 9
∴ k = 9/-9
∴ k = -1
If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2 = 1 + 4q.
4. If the roots
of the equation x2 + px + q = 0 differ by 1, prove that p2
= 1 + 4q.
Sol. x2 + px + q = 0
Comparing with ax2 + bx + c = 0 we have
a = 1, b = p, c = q
Let α
and β be the roots
of given quadratic equation.
α –
β = 1 ......(i) [Given]
α + β =
-b/a = -p/1 = -p
Also, α . β = c/a = q/1
= q
We know that,
(α – β )2 = (α + β )2 – 4αβ
∴ (1)2 = (– p)2 – 4 (q)
∴ 1 = p2 – 4q
∴ 1 + 4q = p2
∴ p2 = 1 + 4q
Hence proved.
Find k, if one of the roots of the quadratic equation kx2 – 7x + 12 = 0 is 3.
3. Find k,
if one of the roots of the quadratic equation kx2 – 7x + 12 =
0 is 3.
Sol. kx2 – 7x + 12 = 0
∵ 3 is one of the root of the quadratic equation, kx2 – 7x + 12 = 0
∴ x = 3
satisfies the equation,
Substituting x = 3 in equation we get,
k (3)2 – 7 (3) + 12 = 0
∴ 9k – 21 + 12 = 0
∴ 9k – 9 = 0
∴ 9k = 9
∴ k = 9/9
∴ k = 1
Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in the ratio 2 : 5.
2. Find k,
if the roots of the quadratic equation x2 + kx + 40 = 0 are in the ratio 2 : 5.
Sol. x2 + kx + 40 = 0
Comparing with ax2 + bx + c =
0 we have a = 1, b = k, c = 40
Let α and β be the roots of given quadratic equation.
Ratio of α to
β is 2 : 5 [Given]
Let the common multiple be m
∴ α = 2m and β = 5m
∴ α +β = - b/a = -k/1 = - k
∴ 2m + 5m = - k
∴ 7m = -k
∴ k = -7m
Also, α .β = c/a = 40/1 = 40
∴ 2m× 5m = 40
∴ 10m2 = 40
∴ m2 = 40/10
∴ m2 = 4
∴ m = ±√4
∴ m = ±2
But, k = -7m
∴ k = -7(2)
or -7(-2)
∴ k = -14 or 14
If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the other, find k.
1. If one root
of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the
other, find k.
Sol. kx2 – 5x + 2 = 0
Comparing with ax2 + bx + c =
0 we have a = k, b = – 5, c = 2
Let α and β be the roots of given quadratic equation.
α = 4 β [Given]
∴ α +β = - b/a = - (- 5)/k
= 5/k
∴ 4β + β = 5/k
∴ 5β = 5/k
∴ β = 5/5k
∴ β = 1/k
Also, α.β =
c/a = 2/k
∴ 4β .β = 2/k
∴ 4(1/k)(1/k) = 2/k
∴ 4/k2 =
2/k
∴ 4/2 = k2/k
∴ 2 = k
k2x2 – 2 (k – 1)x + 4 = 0
(ii) k2x2 – 2 (k – 1)x + 4 = 0
Sol. k2x2 – 2 (k – 1)x + 4 = 0
Comparing with ax2 + bx + c = 0 we have a = k2, b = – 2 (k – 1), c = 4
We know that,
∆ = b2 – 4ac
= [– 2 (k – 1)]2 – 4 (k2) (4)
= (– 2k + 2)2 – 16k2
= 4k2 – 8k + 4 – 16k2
= – 12k2 – 8k + 4
∵ The roots of given equation are real and equal.
∴ ∆ must be zero.
∴ – 12k2 – 8k + 4 = 0
∴ – 4 (3k2 + 2k – 1) = 0
∴ 3k2 + 3k – k – 1 = -0/4
∴ 3k (k + 1) – 1 (k + 1) = 0
∴ (k + 1) (3k – 1) = 0
∴ k + 1 = 0 or 3k – 1 = 0
∴ k = – 1 or 3k = 1
∴ k = -1 or k = 1/3
(k – 12)x2 + 2 (k – 12)x + 2 = 0
(i) (k – 12)x2 + 2 (k – 12)x + 2 = 0
Sol. (k – 12)x2 + 2 (k – 12)x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = k – 12, b = 2 (k – 12), c = 2
We know that,
∆ = b2 – 4ac
= [2 (k – 12)]2 – 4 (k -12) (2)
= (2k – 24)2 – 8 (k – 12)
= 4k2 – 96k + 576 – 8k + 96
= 4k2 – 104k + 672
∵ The roots of given equation are real and equal.
∴ ∆ must be zero.
∴4k2 – 104k + 672 = 0
∴4 (k2 – 26k + 168) = 0
∴ k2 – 14k – 12k + 168 = 0
∴ k (k – 14) – 12 (k – 14)= 0
∴ (k – 14) (k – 12) = 0
∴k – 14 = 0 or k – 12 = 0
∴ k = 14 or k = 12
2x2 + 5√3 x + 16 = 0
(vi) 2x2 + 5√3 x + 16 = 0
Sol. 2x2 + 5√3 x + 16 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = 5 3 , c = 16
∴ ∆ = b2 – 4ac
= (5√3 )2 – 4 (2) (16)
= 25 × 3 – 128
= 75 – 128
= – 53
∴ ∆ < 0
Hence roots of the quadratic equation are not real.
2y2 + 5y – 3 = 0
(iv) 2y2 + 5y – 3 = 0
Sol. 2y2 + 5y – 3 = 0
Comparing with ay2 + by + c = 0 we have a = 2, b = 5, c = – 3
∆ = b2 – 4ac
= (5)2 – 4 (2) (– 3)
= 25 + 24
= 49
∴ ∆ > 0
Hence roots of the quadratic equation are real and unequal.
3y2 + 9y + 4 = 0
(v) 3y2 + 9y + 4 = 0
Sol. 3y2 + 9y + 4 = 0
Comparing with ay2 + by + c = 0 we have a = 3, b = 9, c = 4
∆ = b2 – 4ac
= (9)2 – 4 (3) (4)
= 81 – 48
= 33
∴ ∆ > 0
Hence roots of the quadratic equation are real and unequal.
y2 + 8y + 4 = 0
(iii) y2 + 8y + 4 = 0
Sol. y2 + 8y + 4 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 8, c = 4
∆ = b2 – 4ac
= (8)2 – 4 (1) (4)
= 64 -16
= 48
∴ ∆ > 0
Hence roots of the quadratic equation are real and unequal.
y2 + 6y – 2 = 0
(ii) y2 + 6y – 2 = 0
Sol. y2 + 6y – 2 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 6, c = – 2
∆ = b2 – 4ac
= (6)2 – 4 (1) (– 2)
= 36 + 8
= 44
∆ > 0
Hence roots of the quadratic equation are real and unequal.
y2 – 4y – 1 = 0
(i) y2 – 4y – 1 = 0
Sol. y2 – 4y – 1 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = – 4, c = – 1
∆ = b2 – 4ac
= (– 4)2 – 4 (1) (-1)
= 16 + 4
= 20
∴ ∆ > 0
Hence roots of the quadratic equation are real and unequal.
x2 + 4x + k = 0
(vi) x2 + 4x + k = 0
Sol. x2 + 4x + k = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = k
∆ = b2 – 4ac
= (4)2 – 4 (1) (k)
= 16 – 4k
∴ ∆ = 16 – 4k
4x2 + kx + 2 = 0
(v) 4x2 + kx + 2 = 0
Sol. 4x2 + kx + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = k, c = 2
∆ = b2 – 4ac
= (k)2 – 4 (4) (2)
= k2 – 32
∴ ∆ = k2 – 32
√3 x2 + 2√2 x – 2√3 = 0
(iv) √3 x2 + 2√2 x – 2√3 = 0
Sol. √3 x2 + 2√2 x – 2√3 = 0
Comparing with ax2 + bx + c = 0 we have a = √3 , b = 2√2 , c = –2√3
∆ = b2 – 4ac
= (2√2)2 – 4 (√3) (–2√ 3)
= (4 × 2) + (8 × 3)
= 8 + 24
= 32
∴ ∆ = 32
x2 + x + 1 = 0
(iii) x2 + x + 1 = 0
Sol. x2 + x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 1, c = 1
∆ = b2 – 4ac
= (1)2 – 4 (1) (1)
= 1 – 4
= – 3
∴ ∆ = – 3
3x2 + 2x – 1 = 0
(ii) 3x2 + 2x – 1 = 0
Sol. 3x2 + 2x – 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 3, b = 2, c = – 1
∆ = b2 – 4ac
= (2)2 – 4 (3) (– 1)
= 4 + 12
= 16
∴ ∆ = 16
x2 + 4x + 1 = 0
(i) x2 + 4x + 1 = 0
Sol. x2 + 4x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = 1
∆ = b2 – 4ac
= (4)2 – 4 (1) (1)
= 16 – 4
= 12
∴ ∆ = 12
4x2 + 7x + 2 = 0
(xii) 4x2 + 7x + 2 = 0
Sol. 4x2 + 7x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = 7, c = 2
b2 – 4ac = (7)2 – 4 (4) (2)
= 49 – 32
= 17
By Formula method,
x
|
=
|
- b ± √(b2 – 4ac)
| ||
2a
| ||||
∴ x
|
=
|
-(7) ± √17
| ||
2(4)
| ||||
∴ x
|
=
|
-7 ±√17
| ||
8
| ||||
∴ x
|
=
|
-7 + √17
|
or
|
-7 - √17
|
8
|
8
| |||
3q2 = 2q + 8
(xi) 3q2 = 2q + 8
Sol. 3q2 = 2q + 8
3q2 – 2q – 8 = 0
Comparing with aq2 + bq + c = 0 we have a = 3, b = – 2, c = – 8
b2 – 4ac = (– 2)2 – 4 (3) (– 8)
= 4 + 96
= 100
By Formula method,
q
|
=
|
- b ± √(b2 – 4ac)
| ||
2a
| ||||
∴ q
|
=
|
-(-2) ± √100
| ||
2(3)
| ||||
∴ q
|
=
|
2 ±10
| ||
6
| ||||
∴ q
|
=
|
2 ± 10
| ||
6
| ||||
∴ q
|
=
|
2 + 10
|
or
|
2 – 10
|
6
|
6
| |||
∴ q
|
=
|
12
|
or
|
-8
|
6
|
6
| |||
∴ q
|
=
|
2
|
or
|
-4/3
|
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