How do you solve using the completing the square method $x^{2} + 2 x - 63 = 0$ $x^2 + 2x - 63 = 0$ ?

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How do you solve using the completing the square method $x^{2} + 2 x - 63 = 0$ $x^2 + 2x - 63 = 0$ ?

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Answer 1 · 2016-03-08T04:28:59

$(x + 9) (x - 7)$ $(x+9)(x-7)$

Explanation:

To solve the equation $x^{2} + 2 x - 63 = 0$ $x^2+2x−63=0$ , using the completing the square method, one has to complete the square using the variable $x$ $x$ .

We know the identity ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ , hence we should halve the coefficient of $x$ $x$ and square it and then add and subtract the 'square' in the trinomial. Here coefficient of $x$ $x$ is 2 and hence squareof half of it is $1$ $1$ . Hence

$x^{2} + 2 x - 63 = 0$ $x^2+2x−63=0$ can be written as

$x^{2} + 2 x + 1 - 1 - 63 = 0$ $x^2+2x+1-1−63=0$ or

$(x^{2} + 2 \times x \times 1 + 1^{2}) - 64$ $(x^2+2xxx xx1+1^2)-64$ or

${(x + 1)}^{2} - 8^{2}$ $(x+1)^2-8^2$

Now using the identity $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ , this becomes

$((x + 1) + 8) \times ((x + 1) - 8)$ $((x+1)+8)xx((x+1)-8)$ or

$(x + 9) (x - 7)$ $(x+9)(x-7)$

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How do you solve using the completing the square method $x^{2} + 2 x - 63 = 0$ $x^2 + 2x - 63 = 0$ ?

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Explanation:

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How do you solve using the completing the square method x^2 + 2x - 63 = 0x2+2x−63=0x^2 + 2x - 63 = 0?

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How do you solve using the completing the square method $x^{2} + 2 x - 63 = 0$ $x^2 + 2x - 63 = 0$ ?