# KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers

## KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers

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3x^{2}- 5x+k=0 are equal.

Here a = 3 , b = -5 , c = k

Since roots are equal then discriminate is equal to zero. i.e.

B^{2}– 4ac=0 ………..(ii)put values in (ii)

=> (-5)^{2} – 4 (3)(k) = 0 => 25 = 12k

=> k = 25/12

Evaluate ( -1+ √(-3) )^{7}+ ( -1- √(-3) )^{7}.

Solution: ( -1+ √(-3) )^{7}+ ( -1- √(-3) )^{7}

As (-1+i√3)/2 = w and (-1+i√3)/2 =w^{2}

=> -1+i √3 = 2w and -1-i√3 = 2w^{2}

Then

( -1+ √(-3) )^{7}+ ( -1- √(-3) )^{7} = (2w)^{7} + (2w^{2} )^{7}

=> 2^{7} w^{7}+2^{7} w^{14}

=> 2^{7} (w^{2}.w+w^{12}.w^{2})

=> 2^{7} [(w^{3} )^{2}.w+(w^{3} )^{4}.w^{2}]

As w^{3} = 1

=> 2^7 ((1)^{3}.w+(1^{4}.w^{2})

=> 2^{7} (w+w^{2})

=> 128 (-1) AS w+ w^{2} = -1

=> -128

(i) 4x^{2}- 1=0 (ii) 3x^{2}+ 4x =0

Solution: (i) 4x^{2}+ 0.x – 1 =0

Here a = 4 , b = 0 , c = -1

As sum of the roots S = – b/a……………(i)

Put values (i).

S = – 0/4 => S = 0

Now product of the roots = p = S = c/a …………….. (ii) put values in ……(i)

=> P = – 1/4

(ii) 3x^{2}+ 4x+0 = 0

Here a = 3 , b = 4 , c = 0

As sum of the roots S = – b/a …………(i)

Put values in (i).

S = – 4/3

Product of the roots = S = – c/a put values

=> P = – 0/3 => P = 0

Solution: 3x^{2}+ (2K+1 )x+k-5 = 0……………(i)

Here a = 3 , b = 2k + 1 , c k – 5

As sum of the roots S = – b/a …………..(i)

Put values in (i).

=> S = – (2k+1)/3 …………………. (ii)

As product of the roots = P = – c/a ………………… (iii) put values in (iii)

P = (k-5)/3…………….. (iv)

AS Sum of the roots = product of the roots

Then from (ii) and (iv)

– (2k+1)/3 = (k – 5)/3 => -(2k + 1 ) = k – 5

=> – 2k – 2 = k – 5 => – 2 + 5 = k + 2k

=> 3k = 3 => K = 3/3 = 1

^{2}- 3x+k+1=0 differ by unity.

Solution: x^{2}– 3x+k+1=0……………(i)

Let a and β are the roots of equation ……(i)

α-β = 1 ……………….. (i)

Here a = 1 , b = -3 , c = k + 1

Sum of the roots α+β= -b/a

α+β-(-3)/1 => α+β=3 ……………………….(ii)

Product of the roots αβ= c/a

αβ= (k+1)/1 => αβ= k+1…………….(iii)

Adding (i) and (ii)

α-β= 1

= (α+β=3)/(2a=4)

=> α=4/2 => α = 2

Put α=2 in (ii)

2 + β = 3 => β =3-2

=> β = 1

Now put α=2 and β = 1 in ( iii)

(2)(1) = k + 1 => 2 – 1 = k

=> k = 1

Thus k = 1