Help me with discriminant

Question

If the equation (a+4)x^2-(3a-1)x+25=0, where a is not equal to -4, has one double real root, find the values of a

Answer 1

For a quadratic equation set equal to zero as in ax²+bx+c = 0, the discriminant is defined as b²-4ac. When the value of the discriminant is zero, the equation has one real double root. If a, b, c are integers (a≠0), then the double root is rational.

Writing your equation using the letter k (instead of a):

(k+4)x²-(3k-1)x+25 = 0, it is then the case that a = k+4, b=-3k+1, c = 25.

An expression for the value of the discriminant is then:

(-3k+1)²-4(k+4)(25)

Simplifying and setting equal to 0:

9k²-6k+1-100k-400 = 0

9k²-106k-399= 0

That is another quadratic equation which can be solved by factoring or using the quadratic formula.

9x²+27x-133x-399=0 (split the middle)

9x(x+3)-133(x+3)=0 (factor by grouping)

(9x-133) (x+3)=0

x= 133/9 or x = -3

Those are the values of k (or a in the original problem) that will result in a solution that has exactly one double root.

For example when k=-3

x²+10x+25=0

x=-5 (multiplicity 2)

Left to you to try k = 133/9.