Question

A quadratic equation of the form 0 = ax2 + bx + c has a discriminant value of –16. How many real number solutions does the equation have?

Answer 1

Solution: The equation $ax^2+bx+c=0$ has zero real number solutions if the value of the discriminant is -16.

Explanation:

The given quadratic equation is $ax^2+bx+c=0$. Since it is quadratic equation therefore it has at two possible solutions either real or imaginary.

To find the solution we use the quadratic formula.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The value $b^2-4ac$ is known as discriminant. Since the discriminant is in the square root, therefore the value of discriminant must be positive to get a real solution.

Here are three cases to decide the nature f solution.

If $b^2-4ac>0$, then the quadratic equation has two real solutions.

If $b^2-4ac=0$, then the quadratic equation has real solution, i.e., $\frac{-b}{2a}$.

If $b^2-4ac<0$, then the quadratic equation has no real solution because $\sqrt{-(b^2-4ac)}$ is an imaginary number.

It the given equation the value of discriminant is -16, which is a negative value, so the given quadratic equation has 2 imaginary solutions.

Therefore, the equation $ax^2+bx+c=0$ has zero real number solutions if the value of the discriminant is -16.

Answer 2

Answer:

B. 0

Step-by-step explanation:

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