Q: How do you solve 8x2 plus 16x plus 8 equals 0?

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8x2 + 16x + 8 = 8*(x2 + 2x + 1) = 8*(x + 1)2

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x=5

the roots are 1, - 1/3, - 3 1/2 and 3 1/2

You could try dividing both numerator and denominator by 8x2 .

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8x2 + 16x + 8 = 8*(x2 + 2x + 1) = 8*(x + 1)2

Please do not remove this question from Inappropriate or split any alts from it. Thanks!

24x4 + 16x² - 8 = 8(3x4 + 2x² - 1) = 8(x² + 1)(3x² - 1).

x=5

15 = 8x2 - 14x; whence, 8x2 - 14x - 15 = (2x - 5)(4x + 3) = 0. Therefore, x = 5/2 or -3/4.

16.

It is: 25-4(8*6) = -167

the roots are 1, - 1/3, - 3 1/2 and 3 1/2

(4x + 3)(2x - 1) so x = -0.75 or 0.5

You could try dividing both numerator and denominator by 8x2 .

8x2 DUH

x3 + 7x = 8x2x3 + 7x - 8x2 = 0 [subtract 8x2 from both sides]x(x2 + 7 - 8x) = 0 [factor out x]x(x-1)(x-7) = 0 [factor]Since the product of the three factors x, x-1, and x-7 equals zero, any of the three expressions could equal zero:x = 0x - 1 = 0, x = 1x - 7 = 0, x = 7Therefore, there are three solutions to the equation x3 + 7x = 8x2:x = 0x = 1x = 7Or, in set notation: x = {0,1,7}