which value of b will cause the quadratic equation x² + bx + 5 = 0 to have two real number…

Answer:

A. -5

Explanation:

Step1: Recall the discriminant formula

For a quadratic equation $ax^{2}+bx + c=0$, the discriminant $\Delta=b^{2}-4ac$. Here $a = 1$, $c = 5$.

Step2: Set up the condition for real - solutions

A quadratic equation has two real number solutions when $\Delta>0$. So we need $b^{2}-4(1)(5)>0$, which simplifies to $b^{2}-20>0$.

Step3: Test each option

• For $b=-5$: $(-5)^{2}-20=25 - 20=5>0$.
• For $b = - 3$: $(-3)^{2}-20=9 - 20=-11<0$.
• For $b = 3$: $3^{2}-20=9 - 20=-11<0$.
• For $b = 4$: $4^{2}-20=16 - 20=-4<0$.

So the value of $b$ that causes the quadratic equation to have two real - number solutions is $b=-5$.