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


x2+6x+10=0



{−3+i, −3−i}


{−3+2i, −3−2i}


{−6+2i, −6−2i}


{−2i, −4i}



Jamal solved the equation by completing the square.


x2−8x+14=0




Which equation shows one of the steps Jamal could have taken to complete the square?





x2−8x+16=−14+16


x2−8x+64=−14+64


x2−8x+16=−14


x2−8x+64=−14

Respuesta :

(1)

we are given

[tex]x^2+6x+10=0[/tex]

we can use quadratic formula

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

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

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

we get

[tex]a=1,b=6,c=10[/tex]

now, we can plug these values into quadratic formula

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

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

we can simplify it

[tex]x=-3+i[/tex]

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

[tex]x=-3-i[/tex]

so, we will get solution

{−3+i, −3−i}.........Answer

(2)

we are given equation as

[tex]x^2-8x+14=0[/tex]

Since, Jamal solve this equation by completing square

so, firstly we will move constant term on right side

so,  subtract both sides by 14

[tex]x^2-8x+14-14=0-14[/tex]

[tex]x^2-8x=-14[/tex]

we can write

[tex]-8x=-2\times 4\times x[/tex]

so, we will add both sides by 4^2

[tex]x^2-8x+4^2=-14+4^2[/tex]

we get

[tex]x^2-8x+16=-14+16[/tex]..............Answer


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