Question

Which value of b will cause the quadratic equation x^2+bx+5=0 to have two real number solution

Answer 1

Answer:

Every value of b>4.47 and b<-4.47 will cause the quadratic equation $x^2+bx+5=0$ to have two real number solution.

Step-by-step explanation:

We have the quadratic function $x^2+bx+5=0$, and we have to find the value of b.

A quadratic function is $ax^2+bx+c=0, a\neq 0$, a quadratic function usually has two real solutions. You can find that solutions using Bhaskara's Formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

$x_1$ and $x_2$ are real solutions of the quadratic equation if and only if:

$b^2-4.a.c >0$

If $b^2-4.a.c$ the quadratic equation doesn't have real solutions.

If  $b^2-4.a.c =0$  the quadratic equation has only one solution.

Then in this case to have two real number solutions: $b^2-4.a.c >0$

We have $x^2+bx+5=0$, where a=1, b, c=5

Then,

$b^2-4.a.c >0\\b^2-4.1.5>0\\b^2-20>0$

Adding 20 in both sides of the equation:

$b^2-20>0\\b^2-20+20>20\\b^2>20\\b>\sqrt{20}$

Which is the same as: $b$

Then, $b>\sqrt{20}\\b>4.47\\b$

Then every value of b>4.47 and b<-4.47 will cause the quadratic equation $x^2+bx+5=0$ to have two real number solution.

For example b=-5 or b=5.

If you replace with b=-5 in $b^2-4.a.c >0$

$b^2-4.a.c >0\\(-5)^2-4.1.5>0\\25-20>0\\5>0$

Then the quadratic function has two real number solutions.

Answer 2

To have 2 real number solutions  the discriminant b^2 - 4ac  must be greater than zero.

so the condition  for 2 real roots in this equation is:-

b^2 - 4*1*5 > 0
b^2 >  20

b > sqrt 20    ( positive root)    Answer