Here we are going to see some practice questions on quadratic functions.

Question 1 :

If the difference of the roots of the equation 2x2 − (a + 1)x + a −1 = 0 is equal to their product, then prove that a = 2.

Solution :

Let α and β be two roots of the given quadratic equation,

Given that :

α - β = α β

From the given quadratic equation, let us find sum and product of roots.

2x2 − (a + 1)x + a −1 = 0

 α + β = -b/aα + β = (a + 1)/2 α β = c/aα β = (a - 1)/2

Formula for α - β = √(α + β)2 - 4αβ

√(α + β)2 - 4αβ  =  αβ

√((a+1)/2)2 - 4(a-1)/2  =  (a - 1)/2

√((a+1)/2)2 - 2(a-1)  =  (a-1)/2

√((a2 + 2a + 1) - 8(a-1))/2  =  (a-1)/2

√[(a2+2a+1)-8a+8]  =  a-1

√(a- 6a + 9)  =  a - 1

Taking squares on both side, we get

a- 6a + 9  =  (a - 1)2

a- 6a + 9  =  a2 - 2a + 1

-6a + 2a  =  1 - 9

-4a  =  -8

a  =  2

Hence proved.

Question 2 :

Find the condition that one of the roots of ax2 + bx + c may be (i) negative of the other, (ii) thrice the other, (iii) reciprocal of the other.

Solution :

ax2 + bx + c

Let α and β be the roots of a quadratic equation

(i)  α  =  -β (one root is negative of other), β  =  β

Sum of roots  =  α  + β  =  -b/a

-β + β  =  -b/a

0  =  -b/a

b = 0 is the required condition.

(ii)  α  =  3β (one root is thrice the other), β  =  β

 Sum of roots (α+β) =  -b/a3β + β  =  -b/a4β  =  -b/aβ  =  -b/4a  --(1) Product of roots αβ = c/a3β + β  =  c/a3β2  =  c/a  ---(2)

Applying the value of β int the 2nd equation, we get

3(-b/4a)2  =  c/a

3b2/16a2  =  c/a

3b=  16a2c/a

3b2  =  16ac

(iii) reciprocal of the other

α  =  1/β (one root is reciprocal of other), β  =  β

Sum of roots (α+β) =  -b/a

1/β + β  =  -b/a

Product of roots (αβ)=c/a

(1/β) ⋅ β  =  c/a

1  =  c/a

a  =  c is the required condition.

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