Question

Does every quadratic equation have at

Answer 1

Answer:

D. No. When the discriminant is less than zero, there are no real solutions.

Step-by-step explanation:

We have to notice that quadratic equations are second order polynomials whose standard form is:

$y = a\cdot x^{2}+b\cdot x + c \cdot x$, $\forall \,a,b,c\,\in\,\mathbb{R}$

The factorized form of this polynomial is:

$y = (x-r_{1})\cdot (x-r_{2})$

Where $r_{1}$ and $r_{2}$ are the roots of the quadratic equation, which may be real or complex.

If we equalize $y$ to zero and make algebraic handling, we can get the value of each root anatically by the Quadratic Formula:

$r_{1,2} = \frac{-b\pm \sqrt{b^{2}-4\cdot a \cdot c}}{2\cdot a}$

Where $b^{2}-4\cdot a\cdot c$ is known as the discriminant. Then, we should remember the following rules:

i) If discriminant is greater than zero, then the quadratic equation has two different real roots.

ii) If discriminant is zero, then the quadratic equation has two equal real roots.

iii) If discriminant is less than zero, then the quadratic function has two complex roots.

In consequence, we came to the conclusion that statement is false, as quadratic equation have all roots real or all complex, but not one real and the other complex.

In a nutshell, the correct answer is D.