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8x^2+16x+8Find the real/complex root of 8x^2+16+8

User Eric Bronnimann
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1 Answer

14 votes
14 votes

Solution:

Consider the polynomial:


8x^2+16x+8

now, to find the roots of the above equation, use the following quadratic equation:


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

in this case, we have that a = 8, b = 16 and c = 8. So, replacing the above data on the previous equation we get:


x=\frac{-16\pm\sqrt[]{16^2^{}-4(8)(8)}}{2(8)}

this gives us :


x\text{ = }-1

thus, we can conclude that the root of the given polynomial is


x\text{ = }-1

User LPL
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