How do you solve $x^{2} - 12 x + 32 = 0$ $x^2-12x+32=0$ ?

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How do you solve $x^{2} - 12 x + 32 = 0$ $x^2-12x+32=0$ ?

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Answer 1 · 2016-06-22T18:03:36

$x = 4$ $x=4$ or $x = 8$ $x=8$

Explanation:

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Method 1 - Finding a pair of factors

Find a pair of factors of $32$ $32$ with sum $12$ $12$ .

The pair $4, 8$ $4, 8$ works in that $4 \times 8 = 32$ $4xx8 = 32$ and $4 + 8 = 12$ $4+8=12$

Hence we find:

$0 = x^{2} - 12 x + 32 = (x - 4) (x - 8)$ $0 = x^2-12x+32 = (x-4)(x-8)$

So $x = 4$ $x = 4$ or $x = 8$ $x = 8$

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Method 2 - Completing the square

$0 = x^{2} - 12 x + 32$ $0 = x^2-12x+32$

$= x^{2} - 2 (6 x) + 36 - 4$ $= x^2-2(6x)+36-4$

$= {(x - 6)}^{2} - 2^{2}$ $= (x-6)^2-2^2$

$= ((x - 6) - 2) ((x - 6) + 2)$ $= ((x-6)-2)((x-6)+2)$

$= (x - 8) (x - 4)$ $= (x-8)(x-4)$

Hence $x = 8$ $x=8$ or $x = 4$ $x=4$

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Method 3 - Quadratic formula

$x^{2} - 12 x + 32 = 0$ $x^2-12x+32 = 0$

is in the form $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ with $a = 1$ $a=1$ , $b = - 12$ $b=-12$ , $c = 32$ $c=32$

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$= \frac{12 \pm \sqrt{12^{2} - 4 (1) (32)}}{2 \cdot 1}$ $= (12+-sqrt(12^2-4(1)(32)))/(2*1)$

$= \frac{12 \pm \sqrt{144 - 128}}{2}$ $= (12+-sqrt(144-128))/2$

$= \frac{12 \pm \sqrt{16}}{2}$ $= (12+-sqrt(16))/2$

$= \frac{12 \pm 4}{2}$ $= (12+-4)/2$

$= 6 \pm 2$ $= 6+-2$

So $x = 8$ $x=8$ or $x = 4$ $x=4$

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How do you solve $x^{2} - 12 x + 32 = 0$ $x^2-12x+32=0$ ?

1 Answer

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