2x^2+8x+8=0 25e^2+90e+81=0 The answers should be in x=? x=?
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2x^2+8x+8 = 0 2(x^2+4x+4) = 0 x^2+4x+4 factors into (x+2)(x+2) 2(x+2)^2= 0 X = -2 I don't know about the second one.
Originally Posted by Belanova 2x^2+8x+8=0 25e^2+90e+81=0 The answers should be in x=? x=? In general, a^2 + 2ab + b^2 = (a + b)^2. So: 2(x^2 + 4x + 4) = 0 => 2(x + 2)^2 = 0 ....... (5e + 9)^2 = 0 ........
Thanks very much. What about these? 48q^2+168q+147=0 147r^2+126r+27=0
Originally Posted by Belanova Thanks very much. What about these? 48q^2+168q+147=0 Mr F says: 3(4q + 7)^2 = 0 147r^2+126r+27=0 Mr F says: Take out a common factor (hint: 3) and write as a perfect square. Same as the others. ..
Wow, that makes those problems incredibly simple. I was never taught that, thanks. Originally Posted by mr fantastic In general, a^2 + 2ab + b^2 = (a + b)^2. So: 2(x^2 + 4x + 4) = 0 => 2(x + 2)^2 = 0 ....... (5e + 9)^2 = 0 ........
Originally Posted by coach2uf Wow, that makes those problems incredibly simple. I was never taught that, thanks. It only works for a quadratic trinomial that's a perfect square. That is, something that has the form $\displaystyle a^2 \pm 2ab + b^2$.
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