# For what values of a does the quadratic equation ax^2 + 20x + 4 = 0 have equal roots?

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### 2 Answers

The roots of a quadratic equation ax^2 + bx + c = 0 can be represented as [–b + sqrt (b^2 – 4ac)]/2a and [–b - sqrt (b^2 – 4ac)]/2a

Now if the two are equal [–b + sqrt (b^2 – 4ac)]/2a = [–b - sqrt (b^2 – 4ac)]/2a

cancel 2a

=> [–b + sqrt (b^2 – 4ac)] = [–b - sqrt (b^2 – 4ac)]

subtract –b

=> sqrt (b^2 – 4ac) = - sqrt (b^2 – 4ac)

=> sqrt (b^2 – 4ac) = 0

=> b^2 – 4ac = 0

Here b = 20 and c = 4

=> 20^2 – 4*a*4 =0

=> 20^2 = 16a

=> a = 20^2 / 16

=> a = 25

**Therefore a is equal to 25.**

To solve for what values the equation ax^2+20x+4 0 has zero roots.

Given that ax^2+20x+4 = 0 has equal roots.

We divide both sides by a. This is possible as a cannot be be zero. Further, if a = 0, then it is a contradicion for ax^2+bx+c to be quadratic.

So x^2+20x/a +4/a is = 0 has equal roots.

Let the equal roots be m. Then x^2+20x/a +c/a = (x-m)^2.

So x^2+20x/a+4/a = x^2-2m+m^2.

So this has to be an identiy.

Sa 20/a = -2m,

or 10/a = -m..(1)

4/a = m^2...(2)

We eliminate m from (1) and (2): m^2 = 4/a = (10/a)^2.

So 4a = 100 . Or a = 100/4 .

a = 25.

Therefore a = 25.