What is true about the solutions of a quadratic equation when the radicand of the quadratic formula is a perfect square?
answer

no real solutions

two identical rational solutions

two different rational solutions

two irrational solutions

The solution of quadratic formula is represented as

x = -b ± squareroot (b^2 – 4ac) / 2a

Where b^2 – 4ac is known as the radicand.

Lets take an equation; x^2 + 6x + 5 = 0

Where a = 1, b = 6, c = 5

The solution will be x = -6 ± squareroot(6^2 – 4(1)(5)) / 2(1)

Where radicand = 6^2 – 4(1)(5) = 36 – 20 = 16

Hence radicand is the perfect square of 4. i.e. 4^2 = 16

Now solving the equation further;

x = (-6 ± 4 ) / 2

x = -1 , x= -5

Thus for a radicand to be a perfect square, the solution of a quadratic equation will composed of two different rational solutions.   

here is the

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