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GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

State quadratic equations have real roots or no real root :- −x2−2x−2=0

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જવાબ : no real roots

State quadratic equations have real roots or no real root :- x−1x−3=0, x≠0

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જવાબ : real roots

State quadratic equations have real roots or no real root :- 2x2−6x+3=0

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જવાબ : real roots

State quadratic equations have real roots or no real root :- −13x2+3x+7=0

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જવાબ : real roots

State quadratic equations have real roots or no real root :- 11x2−12x−1=0

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જવાબ : real roots

State quadratic equations have real roots or no real root :- 9x2+x+3=0

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જવાબ : no real roots

State quadratic equations have real roots or no real root :- 3x2+6x+1=0

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જવાબ : real roots

State quadratic equations have real roots or no real root :- sx2+x+7=0

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જવાબ : no real roots

Find the roots of the quadratic equation using square method :-2x²−7x+3=0

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જવાબ : 2x²−7x+3=0 (x−7/4)2−(7/4)2+3/2=0
(x−7/4)2=49/16−3/2=0
(x−7/4)2=25/16
x−7/4=±5/4
or
x=1/2 or 3

Find the roots of the quadratic equation using square method :- x²+4x−5=0

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જવાબ : (x+4/2)2−(4/2)2−5=0
(x+2)2−9 = 0
(x+2)2= 9
x+2 = ±3
x=1 or -5

Find the roots of the quadratic equation using factorization technique :-

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જવાબ : x²−11x+30=0
Solution- x²−11x+30=0
x(x-5)-(x-6)=0
Roots are 5 and 6

Find the roots of the quadratic equation using factorization technique :-

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જવાબ : x²−3x−10=0
Solution- x²−3x−10=0
x(x−5)+2(x−5)=0
(x+2)(x−5)=0
So roots are x=-2 and 5

Find whether statement is True or False :- Every quadratic equation will have rational roots

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જવાબ : False

Find whether statement is True or False :- if the roots of the quadratic equation are irrational, the coefficient of the term x will also be irrational

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જવાબ : True

Find whether statement is True or False :- if the roots of the quadratic equation are rational, the coefficient of the term x will also be rational.

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જવાબ : True

Find whether statement is True or False :- for k > 0,the quadratic equation 2x2+6x−k=02x2+6x−k=0 will definitely have real roots

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જવાબ : True

Find whether statement is True or False :- In a quadratic equation ax2+bx+c=0ax2+bx+c=0 ,if a and c are of opposite sign, then quadratic equation will definitely have real roots

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જવાબ : True

In a quadratic equation ax2+bx+c=0ax2+bx+c=0 ,if a and c are of same sign and b is zero ,the quadratic equation has real roots

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જવાબ : False

Find whether statement is True or False :- A quadratic equation can have at most 2 real roots

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જવાબ : True

Find whether statement is True or False :- The roots of the equation x2−1=0x2−1=0 are 1,-1

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જવાબ : True

Find whether statement is True or False :- There are no reals roots of the quadratic equation x2+4x+5=0x2+4x+5=0

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જવાબ : True

Find a natural number whose square diminished by 84 is thrice the 8 more of given number

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જવાબ : 12

The roots of the quadratic equation x2+14x+40=0 are

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જવાબ : (-4,-10)

State whether the equation is a quadratic equation or not a quadratic equation x⁵+x+20=0

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જવાબ :
It is not a quadratic equation

The roots of the quadratic equation are real or non-real x2+2x+5=0

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જવાબ : Non-real

Find the roots of 6x²−√2x−2=0 by the factorisation of the corresponding quadratic polynomial.

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જવાબ : √23,√25−23,25

Find Whether the equation is a quadratic equation and isn’t? :- x³−x²+2x+1=(x+1)3

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જવાબ : quadratic equation

Find Whether the equation is a quadratic equation and isn’t? :- (x+2)(x−1)=x²−2x−5

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જવાબ : not quadratic equation

Find Whether the equation is a quadratic equation and isn’t? :- x²+3x=(−1)(1−3x)2

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જવાબ : a quadratic equation

Find Whether the equation is a quadratic equation and isn’t? :- (x+2)2=2(x+3)

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જવાબ : quadratic equation

State the quadratic equations have real roots, no real roots :-2x²−5x+3=0

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જવાબ : y= 3/2 or 1

State the quadratic equations have real roots, no real roots :-3y²+4y=2(y+4)

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જવાબ : y= -2 or 4/3

State the quadratic equations have real roots, no real roots :-14x+4x²=2x−5

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જવાબ : y=-1 or 9/10

State the quadratic equations have real roots, no real roots :-9−y−10y²=0

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જવાબ : y= -1 or 9/10

State the quadratic equations have real roots, no real roots :-11x²−31x−6=0

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જવાબ : x= 3 or -2/11

State the quadratic equations have real roots, no real roots :-y²+(y+2)2=580

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જવાબ : y=16 or -18

State the quadratic equations have real roots, no real roots :-x²+(7−x)2=25

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જવાબ : x= 3 or 4

State the quadratic equations have real roots, no real roots :-2x²−x−2=0

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જવાબ : Real roots

State the quadratic equations have real roots, no real roots :-2x²−6x+3=0

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જવાબ : Real roots

State the quadratic equations have real roots, no real roots :-10x²+3x+7=0

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જવાબ : No real roots

State the quadratic equations have real roots, no real roots :-11x²+5x+1=0

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જવાબ : No real roots

State the quadratic equations have real roots, no real roots :-9x²+x−3=0

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જવાબ : Real roots

State the quadratic equations have real roots, no real roots :-3x²+5x+1=0

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જવાબ : Real roots

State the quadratic equations have real roots, no real roots :- x²+8x+7=0

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જવાબ : Real roots

Find the roots with the quadratic equations :- 6x² – 7x + 2 =0

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જવાબ : 2/3 , ½

Find the roots with the quadratic equations :- x² – 43x +42 = 0

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જવાબ : 1, 42

Find the roots with the quadratic equations :- x² - 5x -6 = 0

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જવાબ : -1, 6

Find the roots with the quadratic equations :- x² + 3x +2=0

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જવાબ : -1, -2

Find the roots with the quadratic equations :- x2 + 6x +5=0

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જવાબ : -5,-1

Had Ram scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

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જવાબ : 15

In the roots of the following quadratic equation: 2√3x² – 5x + √3 = 0 (2011D)

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જવાબ : We have, 2 $\sqrt{3}$ x² – 5x + $\sqrt{3}$ = 0
Here, a = 2 $\sqrt{3}$ , h = -5, c = $\sqrt{3}$
D = b² – 4ac
∴ D = (-5)² – 4 (2 $\sqrt{3}$ )( $\sqrt{3}$ )
= 25 – 24 = 1

Solve for x: 4x² – 4ax + (a² – b²) = 0 (2011OD)

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જવાબ : 4x² – 4ax + (a² – b²) = 0
⇒ [4x² – 4ax + a²] – b² = 0
⇒ [(2x)² – 2(2x)(a) + (a)²] – b² = 0
⇒ (2x – a)² – (b)² = 0
⇒ (2x – a + b) (2x – a – b) = 0
⇒ 2x – a + b = 1 or 2x – a – b = 0
2x = a – b or 2x = a + b
∴ x = $\frac{a-b}{2}$ or x = $\frac{a+b}{2}$

Solve for x: 3x² – 2√6x + 2 = 0 (2012D)

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જવાબ : 3x² – 2 $\sqrt{6}$ x + 2 = 0
⇒ 3x² – $\sqrt{6}$ x – $\sqrt{6}$ x + 2 = 0
⇒ $\sqrt{3}$ x ( $\sqrt{3}$ x – $\sqrt{2}$ – $\sqrt{2}$ ( $\sqrt{3}$ x – $\sqrt{2}$ ) = 0
⇒ ( $\sqrt{3}$ x – $\sqrt{2}$ )( $\sqrt{3}$ x – $\sqrt{2}$ ) = 0
⇒ $\sqrt{3}$ x – $\sqrt{2}$ = 0 ⇒ x = $\frac{\sqrt{2}}{\sqrt{3}}$
∴ x = $\frac{\sqrt{6}}{3}$ ….[ $\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

Solve the equation 14/(x+3) – 1 = 5/(x+1); x ≠ -3, -1, for x. (2014D)

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જવાબ :

⇒ 5(x + 3) = (11 – x) (x + 1)
⇒ 5x + 15 = 11x + 11 – x² – x
⇒ 5x + 15 – 11x – 11 + x² + x = 0
⇒ x² – 5x + 4 = 0
⇒ x² – 4x – x + 4 = 0
⇒ x(x – 4) – 1(x – 4) = 0
⇒ (x – 1) (x – 4) = 0
⇒ x – 1=0 ; x – 4 = 0
⇒ x= 1 ; x = 4

Solve the equation 3/(x+1) – ½ = 2/(3x-1) ; x ≠ -1, x ≠ 1/3 for x. (2014D)

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જવાબ :

⇒ 2(2x + 2) = (5 – x)(3x – 1)
⇒ 4x + 4 = 15x – 5 – 3x² + x
⇒ 4x + 4 – 15x + 5 + 3x² – x = 0
⇒ 3x² – 12x + 9 = 0
⇒ x² – 4x + 3 = 0 …[Dividing the equation by 3
⇒ x² – 3x – x + 3 = 0
⇒ x(x – 3) – 1(x – 3) = 0
⇒ (x – 1) (x – 3) = 0
⇒ x – 1 = 0 or x – 3 = 0
∴ x = 1 ; x = 3

Solve the equation 4/x – 3 = 5/(2x+3) ≠ -1, x ≠ -3/2, for x. (2014D)

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જવાબ : Important Questions for Class 10 Maths Chapter 4 Quadratic Equations 14

⇒ 5x = (2x + 3) (4 – 3x)
⇒ 5x = 8x – 6x² + 12 – 9x
⇒ 5x – 8x + 6x² – 12 + 9x = 0
⇒ 6x² + 6x – 12 = 0
⇒ x² + x – 2 = 0 …[Dividing the equation by 6
⇒ x² + 2x – x – 2 = 0
⇒ x(x + 2) – 1(x + 2) = 0
⇒ x – 1 = 0 or x + 2 = 0
∴ x = 1 or x = -2

If the roots of the quadratic equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c. (2016OD)

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જવાબ : Here ‘a’ = a – b, ‘b’ = b – c, ‘c’ = c – a
As the roots are equal

D = 0
b² – 4ac = 0
⇒ (b – c)² – 4(a – b)(c – a) = 0
⇒ b² + c² – 2bc – 4(ac – a² – bc + ab) = 0
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b2² + c² – 4ab + 2bc – 4ac = 0
⇒ (-2a)² + (b)² + (c)2² + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0
⇒ [(-2a) + (b) + (c)]² = 0 ….[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
Taking square-root on both sides
-2a + b + c = 0
⇒ b + c = 2a ∴ 2a = b + c

Solve for x: √3x² – 2√23x – 2√3 = 0 (2015OD)

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જવાબ : The given quadratic equation can be written as
$\sqrt{3}$ x² – 2 $\sqrt{23}$ x – 2 $\sqrt{3}$ = 0
a = $\sqrt{3}$ , b = -2 $\sqrt{2}$ , c= -2 $\sqrt{3}$
D = b² – 4ac

Solve for x: 2x² + 6√3x – 60 = 0 (2015OD)

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જવાબ : Given equation can be written as
2(x² + 3 $\sqrt{3}$ x – 30) = 0
x² + 3 $\sqrt{3}$ x – 30 = 0

, a = 1, b = 3 $\sqrt{3}$ , C = -30
D = b² – 4ac

Find that value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3 (p + 9) = 0, p ≠ -1 had equal roots. (2015D)

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જવાબ : For the given quadratic equation to have equal roots, D = 0
a = (p + 1), b = -6(p + 1), c = 3(p + 9)
D = b² – 4ac
⇒ [-6(p + 1)]² – 4(p + 1).3 (p + 9) = 0
⇒ 36(p + 1)² – 12(p + 1) (p + 9) = 0
⇒ 12(p + 1) (3p + 3 – p – 9) = 0
⇒ 12(p + 1)(2p – 6) = 0
⇒ 24(p + 1)(p – 3) = 0
⇒ p + 1 = 0 or p – 3 = 0
⇒ p = -1 (rejected) or p = 3
∴ p = 3

Find that non-zero value of k, for which the quadratic equation kx² + 1 – 2(k – 1)x + x² = 0 has equal roots. Hence find the roots of the equation. (2015D)

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જવાબ : The given quadratic equation can be written as
kx² + x² – 2(k – 1)x + 1 = 0
(k + 1) x² – 2 (k – 1) x + 1 = 0 …(A)
=> a = (k + 1), b = -2(k – 1), c = 1
As the roots are equal, D = 0
D = b² – 4ac
⇒ (-2(k – 1)]² – 4 × (k + 1) × 1 = 0
⇒ 4(k − 1)² – 4(k + 1) = 0
⇒ 4k² + 4 – 8k – 4k – 4 = 0
⇒ 4k² – 12k = 0 ⇒ 4k(k – 3) = 0
k = 3 or k = 0 (rejected)
∴ k = 3
Putting k = 3 put in equation (A), we get
⇒ 4x² – 4x + 1 = 0
⇒ 4x² – 2x – 2x + 1 = 0
⇒ 2x(2x – 1) – 1(2x – 1) = 0
⇒ (2x – 1) (2x – 1) = 0
⇒ 2x – 1 = 0 or 2x – 1 = 0
⇒ x = $\frac{1}{2}$ or x = $\frac{1}{2}$
Roots are $\frac{1}{2} \text { and } \frac{1}{2}$

For what value of k, are the roots of the quadratic equation y² + k² = 2 (k + 1)y equal? (2013OD)

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જવાબ : y² + k² = 2(k + 1)y
y² – 2(k + 1)y + k² = 0
=> a = 1, b = -2(k + 1), c = k²
D = 0 … [Roots are equal
b² – 4ac = 0
∴ [-2(k + 1)]² – 4 × (1) × (k²) = 0
⇒ 4(k² + 2k + 1) – 4k² = 0
⇒ 4k² + 8k + 4 – 4k² = 0
⇒ 8k + 4 = 0
⇒ 8k = -4 ∴ k = $\frac{-4}{8}=\frac{-1}{2}$

For what value of k, are the roots of the quadratic equation: (k – 12)x² + 2(k – 12)x + 2 = 0 equal? (2013OD)

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જવાબ : We have, (k – 12)x² + 2(k – 12)x + 2 = 0
The given quadratic equation will have equal roots if D = 0

⇒ b² – 4ac = 0
=> a = (k – 12), b = 2(k – 12), c = 2
b² – 4ac = 0
0 = 4(k – 12)² – 4 × (k – 12) × 2
0 = (k – 12)[4(k – 12) – 4 × 2]
0 = (k – 12) 4[k – 12 – 2]
0 = 4(k – 12) (k – 14)
∴ 4(k – 12)(k – 14) = 0
∴ k = 12 (rejected) or k = 14
But k cannot be equal to 12 because in that case the given equation will imply 2 = 0 which is not true.
∴ k = 14

For what values of k, the roots of the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 are equal? (2013D)

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જવાબ : We have, (k + 4) x² + (k + 1) x + 1 = 0
=> a = k + 4, b = k + 1, c = 1
As the roots are equal

D =0
b² – 4ac = 0
∴ (k + 1)² – 4(k + 4)(1) = 0
k² + 2k + 1 – 4k – 16 = 0
k² – 2k – 15 = 0
k² – 5k + 3k – 15 = 0
k(k – 5) + 3(k – 5) = 0
(k – 5)(k + 3) = 0
k – 5 = 0 or k + 3= 0
k = 5 or k = -3
∴ k = 5 and -3

For what value of k, the roots of the quadratic equation kx(x – 2√5) + 10 = 0, are equal? (2013D)

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જવાબ : We have, kx(x – 2 $\sqrt{5}$ ) + 10 = 0
kx² – 2 $\sqrt{5}$ kx + 10 = 0
Here a = k, b= -2 $\sqrt{5}$ k, c= 10
As the roots are equal

D = 0
As b² – 4ac = 0
∴ (-2 $\sqrt{5}$ k)² – 4(k)(10) = 0
20k² – 40k = 0
⇒ 20k(k – 2) = 0
∴ 20k = 0 or k – 2 = 0
k = 0 (rejected as Coeff. of x² cannot be zero) or k = 2
∴ k= 2

Find the value of m for which the roots of the equation. mx (6x + 10) + 25 = 0, are equal. (2012OD)

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જવાબ : We have, mx(6x + 10) + 25 = 0
6mx² + 10mx + 25 = 0
Here a = 6m, b = 10m, c = 25
As the roots are equal

D = 0
b² – 4ac = 0
∴ (10m)² – 4(6m) (25) = 0
100m² – 600m = 0 ⇒ 100m (m – 6) = 0
100m = 0 or m – 6 = 0
m = 0 or m = 6
…[Rejecting m = 0, as coeff. of x² cannot be zero
∴ m = 6

Find the value of k for which the roots of the equation kx(3x – 4) + 4 = 0, are equal. (20120D)

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જવાબ : We have, kx(3x – 4) + 4 = 0
3kx² – 4kx + 4 = 0
Here a = 3k, b = -4k, c = 4
As the roots are equal

D = 0
b² – 4ac = 0
∴ (-4k)² – 4(3k) (4) = 0
16k² – 48k = 0
16k (k – 3) = 0
16k = 0 or k – 3 = 0
k = 3 or k = 0…[Rejecting k = 0, as coeff. of x² cannot be zero
∴ k = 3

Solve the following quadratic equation for x: x² – 4ax – b² + 4a² = 0 (2012OD)

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જવાબ : We have, x² – 4ax – b² + 4a² = 0
⇒ x² – 4ax + 4a² – b²= 0
⇒ [x² – 2(x)(2a) + (2a)²] – (b)² = 0
(x – 2a)² – (b)² = 0
(x – 2a + b) (x – 2a – b) = 0
x – 2a + b = 0 or x – 2a – b = 0
∴ x = 2a – b or x = 2a + b

Find the value(s) of k so that the quadratic equation 3x² – 2kx + 12 = 0 has equal roots. (2012D)

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જવાબ : Given: 3x² – 2kx + 12 = 0
Here a = 3, b = -2k, c = 12
As the roots are equal

D = 0
b² – 4ac = 0
∴ (-2k)² – 4(3) (12) = 0
⇒ 4k² – 144 = 0 ⇒ k² = $\frac{144}{4}$ = 36
∴ k = $\pm \sqrt{36}=\pm 6$

Find the value(s) of k so that the quadratic equation 2x² + kx + 3 = 0 has equal roots. (2012D)

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જવાબ : Given: 2x² + kx + 3 = 0
Here a = 2, b = k, c= 3
As the roots are equal

D = 0
As b² – 4ac = 0 ∴ K² – 4(2)(3) = 0
K² – 24 = 0 or k² = 24
∴ k = $\sqrt{2 \times 2 \times 6}=\pm 2 \sqrt{6}$