FINDING UNKNOWN IN THE QUADRATIC EQUATION WITH GIVEN ROOTS

Question 1 :

If one root of the equation

2x2-ax+64  =  0

is twice the other, then find the value of a.

Solution :

Roots of quadratic equation will be α and β.

α  =  2β

By comparing the given equation with general form of quadratic equation, we get

a  =  2, b  =  -a and c = 64

Sum of roots :

α+β  =  -b/a

=  -(-a)/2

α+β  =  a/5  ---(1)

Product of roots :

αβ  =  c/a

= 64/2

αβ  =  32  ---(2)

By applying the value of α in (1), we get

2β+β  =  a/5

3β  =  a/5

By applying the value of α in (2), we get

αβ  =  32

2β (β)  =  32

2  =  32

β  =  √16

β  =  √4 x 4

β  =  4

Then,

3β  =  a/5

3(4)  =  a/5


a  =  60

Question 2 :

If α and β are the roots of

5x2-px+1  =  0

and α - β = 1, then find p.

Solution:

From the given quadratic equation, we get

a = 5, b = -p and c = 1

Sum of the roots :

α+β  =  -b/a

=  -(-p)/5

=  p/5  ----(1)

Product of roots :

αβ  =  c/a

αβ  =  1/5 ----(2)

Given that :

α-β = 1

α-β  =  √(α+β)2-4αβ

By applying the values from (1) and (2), we get

(p/5)2-4(1/5)  =  1

p2/25 - 4/5  =  1

(p2-20)/25  =  1

p2-20  =  25

p2  =  45

p  =  3√5

Question 3 :

If one root of the equation

3x2+kx-81  =  0

is the square of the other, find k.

Solution :

α  =  β2

a = 3, b = k and c = -81

Sum of the roots :

α+β  =  -b/a

α+β  =  -k/3  ---(1)

Product of roots :

 α β  =  c/a

=  -81/2

αβ  =  -27  ---(2)

By applying the value of α in(1), we get

β2+β  =  -k/3 ---- (1)

Applying the value of α in (2), we get

β2β  =  -27

β3  =  (-3)3

β  =  -3

β2+β  =  -k/3

6  =  k/3

18  =  k

So, the value of k is 18.

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