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Jeans123
 3 years ago
solve the quadratic equation by comleting the square,
x^212x+7=0
Jeans123
 3 years ago
solve the quadratic equation by comleting the square, x^212x+7=0

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

Jeans123
 3 years ago
Best ResponseYou've already chosen the best response.0Where did you get 29 from?

ivanmlerner
 3 years ago
Best 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.

yummydum
 3 years ago
Best ResponseYou've already chosen the best response.0k 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
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