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vovikov84 [41]
3 years ago
5

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

Mathematics
1 answer:
vodomira [7]3 years ago
8 0

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

<em>Find the value of D</em>

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

<em>Find out the solutions</em>

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

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