How do you solve using the completing the square method $x^{2} - 8 x - 48 = 0$ $x^2 - 8x -48 = 0$ ?

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How do you solve using the completing the square method $x^{2} - 8 x - 48 = 0$ $x^2 - 8x -48 = 0$ ?

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Answer 1 · 2016-02-28T06:14:24

as follows

Explanation:

$x^{2} - 8 x - 48 = 0$ $x^2-8x-48=0$
$\Rightarrow x^{2} - 2 \cdot x \cdot 4 + 4^{2} - 16 - 48 = 0$ $=>x^2-2*x*4+4^2-16-48=0$
$\Rightarrow x^{2} - 2 \cdot x \cdot 4 + 4^{2} = 64$ $=>x^2-2*x*4+4^2=64$
$\Rightarrow {(x - 4)}^{2} = 64$ $=>(x-4)^2=64$
$\Rightarrow (x - 4) = \pm \sqrt{64} = \pm 8$ $=>(x-4)=+-sqrt64=+-8$
$:.x =8+4=12$ and $x =-8+4=-4$

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How do you solve using the completing the square method $x^{2} - 8 x - 48 = 0$ $x^2 - 8x -48 = 0$ ?

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How do you solve using the completing the square method x^2 - 8x -48 = 0x2−8x−48=0x^2 - 8x -48 = 0?

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How do you solve using the completing the square method $x^{2} - 8 x - 48 = 0$ $x^2 - 8x -48 = 0$ ?