Question

Quincy uses the quadratic formula to solve for the values of x in a quadratic equation. he finds the solution, in simplest radical form, to be x = . which best describes how many real number solutions the equation has? zero, because the discriminant is negative. zero, because the discriminant is not a perfect square. one, because the negative and the minus cancel each other out. two, because of the ± symbol.

Answer 1

The answer is zero because the the discriminant is negative

Answer 2

A quadratic equation in the form of

 a x² + b x + c=0, has

Discriminant = D= b²- 4 ac

x = $\frac{-b \pm \sqrt{D}}{2a}$

Now coming to roots of a quadratic equation

1. D≥0, both the roots are real i.e both of them may be rational or both of them may be rational.

2. D=0, both the roots are real and equal.

3. D< 0, both the roots are imaginary.

So, out of following options given,

Option A, is not true,

zero, because the discriminant is negative.(It is a true statement if you are talking about real roots but if we consider imaginary roots also then this statement becomes false.) .In the question it is given that  that roots are in radical form that's why this option is incorrect.

. Option B is not true because if Discriminant is not a perfect square then also the quadratic function has two real either rational or irrational roots.

Third option is completely false , it is incorrect statement.[one, because the negative and the minus cancel each other]

Fourth option is true because , in the answer it has been written that the roots are in simplest radical form , The value of D should always be greater than zero,then we look at ± symbol .Then there are two possible real roots.