Question

find the discriminant, and determine the number of real solutions. then solve x^2 +8x+20=0

Answer 1

Answer:

Part 1) The quadratic equation has zero real solutions

Part 2) The solutions are

$x_1=-4+2i$   and $x_2=-4-2i$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}+8x+20=0$  

so

$a=1\\b=8\\c=20$

The discriminant is equal to

$D=(b^{2}-4ac)$

If D=0 -----> the quadratic equation has only one real solution

If D>0 -----> the quadratic equation has two real solutions

If D<0 -----> the quadratic equation has two complex solutions

Find the value of D

$D=8^{2}-4(1)(20)=-16$ -----> the quadratic equation has two complex solutions

Find out the solutions

substitute the values of a,b and c in the formula

$x=\frac{-8(+/-)\sqrt{8^{2}-4(1)(20)}} {2(1)}$

$x=\frac{-8(+/-)\sqrt{-16}} {2}$

Remember that

$i=\sqrt{-1}$

$x=\frac{-8(+/-)4i} {2}$

$x_1=\frac{-8(+)4i} {2}=-4+2i$

$x_2=\frac{-8(-)4i} {2}=-4-2i$