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