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

SCIENCE TECHNOLOGY QUESTION PAPERS WITH COMPLETE SOLUTION

SCIENCE & TECHNOLOGY MARCH 2014 PAPER WITH COMPLETE SOLUTION.

SCIENCE PAPER ONE WITH SOLUTION

SCIENCE AND TECHNOLOGY PAPER TWO 

SCIENCE AND TECHNOLOGY PAPER THREE

SCIENCE AND TECHNOLOGY UNIT TEST PAPER

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