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solve the quadratic equation by comleting the square,
x^212x+7=0
 one year ago
 one year ago
solve the quadratic equation by comleting the square, x^212x+7=0
 one year ago
 one year ago

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ivanmlernerBest ResponseYou've already chosen the best response.0
Sum 29 on both sides and you get:\[x^212x+36=(x6)^2=29\]Do square root, isolate x and you get your answer.
 one year ago

Jeans123Best ResponseYou've already chosen the best response.0
Where did you get 29 from?
 one year ago

ivanmlernerBest ResponseYou've already chosen the best response.0
(a+b)^2=a^2+2ab+b^2 To complete the squares is to go from something like the right part to something on the left side. a in this case is clearly x, and because of the second term, we know that b is 6, but we only have 7 on the place of b² and we need to have 36. What is missing for we to get 36 is what we need to add, wich is 29.
 one year ago

yummydumBest ResponseYou've already chosen the best response.0
k i know im late on this but ill explain it anyway... \[x^212x+7=0\]we're gonna complete the square for this first move the \(7\) over to the other side:\[x^212x=7\]now fill in the following blanks with the square of half the middle term [\(({12\over2})^2\)]:\[x^212x+\text{___}=7+\text{___}\]\[x^212x+36=7+36\]now combine like terms:\[x^212x+36=29\]next write the simplified version of \(x^212x+36\) which is \((x6)(x6)\)=\((x6)^2\):\[(x6)^2=29\]square root both sides:\[x6=\sqrt{29}\]add 6 to both sides:\[\large x=6\pm\sqrt{29}\] hope you get how to do this now! :) ...sorry it was so late
 one year ago
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