How do you solve using the completing the square method $x^{2} - 10 x = - 8$ $x^2- 10x = -8$ ?

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Answer 1 · 2018-05-20T17:20:37

$x = 5 \pm \sqrt{17}$ $x = 5+-sqrt17$

$x^{2} - 10 x = - 8$ $x^2- 10x = -8$

$a x^{2} + b x + c$ $ax^2 +bx +c$ .

To complete the square a =1 and $c = {(\frac{1}{2} b)}^{2}$ $c = (1/2b)^2$

$c = {(\frac{1}{2} \cdot - 10)}^{2} = 25$ $c = (1/2*-10)^2 = 25$ , a already is 1

we need to add the c to both sides of the equation so we don't change its value:

$x^{2} - 10 x + c = - 8 + c$ $x^2- 10x + c = -8 + c$

$x^{2} - 10 x + 25 = - 8 + 25$ $x^2- 10x + 25 = -8 + 25$

the -5 is the $\frac{1}{2} b$ $1/2b$ before we squared it:

${(x - 5)}^{2} = 17$ $(x -5)^2 = 17$

$\sqrt{{(x - 5)}^{2}} = \pm \sqrt{17}$ $sqrt((x -5)^2) = +-sqrt17$

$x - 5 = \pm \sqrt{17}$ $x -5 = +-sqrt17$

$x = 5 \pm \sqrt{17}$ $x = 5+-sqrt17$

graph{x^2- 10x +8 [-13.16, 26.84, -18.56, 1.44]}

How do you solve using the completing the square method x2−10x=−8x^2- 10x = -8?