KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers
| Class: | 10th Class | Subject: | Math |
| Chapter: | All | Board: | KPK Boards |
KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers
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For what value of K the roots of the equation</p> <p>3x<sup>2</sup>- 5x+k=0 are equal.
Here a = 3 , b = -5 , c = k
Since roots are equal then discriminate is equal to zero. i.e.
B2– 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 =w2
=> -1+i √3 = 2w and -1-i√3 = 2w2
Then
( -1+ √(-3) )7+ ( -1- √(-3) )7 = (2w)7 + (2w2 )7
=> 27 w7+27 w14
=> 27 (w2.w+w12.w2)
=> 27 [(w3 )2.w+(w3 )4.w2]
As w3 = 1
=> 2^7 ((1)3.w+(14.w2)
=> 27 (w+w2)
=> 128 (-1) AS w+ w2 = -1
=> -128
With out solving the equation, find the sum and products of the roots of the following equation quadratic equation.</p> <p>(i) 4x<sup>2</sup>- 1=0 (ii) 3x<sup>2</sup>+ 4x =0
Solution: (i) 4x2+ 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) 3x2+ 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
Find the value of k so that the sum of the roots of the equation 3x^2+ (2K+1 )x+k-5 = 0 is equal to the product of roots.
Solution: 3x2+ (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
Find the value of k if the roots of x<sup>2</sup>- 3x+k+1=0 differ by unity.
Solution: x2– 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
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