Q:

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

Accepted Solution

A:
Answer:Part 1) The quadratic equation has zero real solutionsPart 2) The solutions are$$x_1=-4+2i$$   and $$x_2=-4-2i$$ Step-by-step explanation:we know thatThe 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 solutionIf D>0 -----> the quadratic equation has two real solutionsIf D<0 -----> the quadratic equation has two complex solutionsFind the value of D$$D=8^{2}-4(1)(20)=-16$$ -----> the quadratic equation has two complex solutionsFind out the solutionssubstitute 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$$i=\sqrt{-1}$$$$x=\frac{-8(+/-)4i} {2}$$ $$x_1=\frac{-8(+)4i} {2}=-4+2i$$ $$x_2=\frac{-8(-)4i} {2}=-4-2i$$