Question

Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?

Answer 1

Answer:

The correct option is 1.

Step-by-step explanation:

The given quadratic equation is

$x^2+bx+5=0$

A quadratic equation $ax^2+bx+c=0$ have to real solution if the value of discriminant is garter than 0.

$b^2-4ac>0$

$b^2-4(1)(5)>0$

$b^2-20>0$

Add 20 on both the sides.

$b^2>20$

It is possible if

$b\sqrt{20}$

$b4.472$

Since -5<-.4.472, therefore option 1 is correct.

Answer 2

Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?


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