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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
+ β = 5/k
∴ 5β = 5/k
∴ β = 5/5k
β = 1/k

Also,  α.β = c/a = 2/k
.β = 2/k
∴ 4(1/k)(1/k) = 2/k
∴ 4/k2 =  2/k
∴ 4/2 = k2/k

∴ 2 = k

SCIENCE TECHNOLOGY QUESTION PAPERS WITH COMPLETE SOLUTION

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 + 22 x – 23 = 0
Comparing with ax2 + bx + c = 0 we have a = 3 , b = 22 , c = –23
∆  = b2 – 4ac
= (22)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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