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

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Which value of b will cause the quadratic equation

Ответ:

isaiyt

13.03.2022

Mathematics

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,

$b^2-4.a.c 0\\b^2-4.1.50\\b^2-200$

Adding 20 in both sides of the equation:

$b^2-200\\b^2-20+2020\\b^220\\b\sqrt{20}$

Which is the same as:

Then, $b\sqrt{20}\\b4.47\\b$

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

$b^2-4.a.c 0\\(-5)^2-4.1.50\\25-200\\50$

Then the quadratic function has two real number solutions.

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

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