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Dmitrij [34]
3 years ago
15

What is the solution set of the equation using the quadratic formula?

Mathematics
1 answer:
oee [108]3 years ago
4 0

(1)

we are given

x^2+6x+10=0

we can use quadratic formula

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

now, we can compare and find a , b and c

we get

a=1,b=6,c=10

now, we can plug these values into quadratic formula

x=\frac{-6\pm \sqrt{6^2-4(1)(10)}}{2(1)}

x=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}

we can simplify it

x=-3+i

x=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}

x=-3-i

so, we will get solution

{−3+i, −3−i}.........Answer

(2)

we are given equation as

x^2-8x+14=0

Since, Jamal solve this equation by completing square

so, firstly we will move constant term on right side

so,  subtract both sides by 14

x^2-8x+14-14=0-14

x^2-8x=-14

we can write

-8x=-2\times 4\times x

so, we will add both sides by 4^2

x^2-8x+4^2=-14+4^2

we get

x^2-8x+16=-14+16..............Answer


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