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Anestetic [448]
3 years ago
15

Find the zeros of x^2+4x=-10 using the quadratic formula

Mathematics
2 answers:
ipn [44]3 years ago
7 0
X^2+4+10=0
x=(-4(+/-)root16-40)/2, so we now know that the zeros are imaginary, because you can't square root a negative number and 16-40 is -24
so the two roots are…
-2+iroot6 and -2-iroot6

Nookie1986 [14]3 years ago
6 0
X² + 4x = 10
x² + 4x - 10 = 10 - 10
x² + 4x - 10 = 0
x = <u>-4 +/- √(4² - 4(1)(-10))</u>
                   2(1)
x = <u>-4 +/- √(16 + 40)</u>
                 2
x = <u>-4 +/ √56
</u>             2<u>
</u>x =<u> -4 +/- √(4 × 14)</u>
               2
x = <u>-4 +/- √4 √14
</u>                2<u>
</u>x = <u>-4 +/- 2√14
</u>              2
x = <u>-4 +/- 2(3.741657387)</u>
                      2
x = <u>-4 +/- 7.483314774</u>
                    2
x = <u>-4 + 7.483314774 </u>   x = <u>-4 - 7.483314774</u>
                   2                                    2
x = <u>3.483314774</u>           x = <u>-11.483314774</u>
               2                                      2
x = 1.741657387           x = -5.741657387

<u />
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Find the axis of symmetry of 2x^2 + 12x + 16 Show your work
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The vertex form:

y=a(x-h)^2+k\\\\(h,\ k)-vertex

The axis of symetry is x = h.

y=ax^2+bx+c\to h=\dfrac{-b}{2a}

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y=2x^2+12x+16\to a=2,\ b=12,\ c=16

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Rus_ich [418]

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<em><u>Solution:</u></em>

From given information,

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