How do you solve $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ by completing the square?

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How do you solve $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ by completing the square?

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Answer 1 · 2018-07-14T19:33:09

$x = - 3 or x = 6$ $x=-3" or "x=6$

Explanation:

$add {(\frac{1}{2} coefficient of the x-term)}^{2} to both sides$ $"add "(1/2"coefficient of the x-term")^2" to both sides"$

$x^{2} + 2 (- \frac{3}{2}) x + \frac{9}{4} = 18 + \frac{9}{4}$ $x^2+2(-3/2)x color(red)(+9/4)=18color(red)(+9/4)$

${(x - \frac{3}{2})}^{2} = \frac{81}{4}$ $(x-3/2)^2=81/4$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x - \frac{3}{2})}^{2}} = \pm \sqrt{\frac{81}{4}} \leftarrow note plus or minus$ $sqrt((x-3/2)^2)=+-sqrt(81/4)larrcolor(blue)"note plus or minus"$

$x - \frac{3}{2} = \pm \frac{9}{2}$ $x-3/2=+-9/2$

$add \frac{3}{2} to both sides$ $"add "3/2" to both sides"$

$x = \frac{3}{2} \pm \frac{9}{2}$ $x=3/2+-9/2$

$x = \frac{3}{2} - \frac{9}{2} = - 3 or x = \frac{3}{2} + \frac{9}{2} = 6$ $x=3/2-9/2=-3" or "x=3/2+9/2=6$

Answer 2 · 2018-07-14T19:38:37

$x = 6 or x = - 3$ $x=6 or x=-3$

Explanation:

Here ,

$x^{2} - 3 x = 18$ $x^2-3x=18$

$=>x^2-3x+k=k+18...to(I)$

We have to find $k$ such that , $x^2-3x+k$
forms a perfect square ,where

$color(blue)(1^(st)term=x^2$ ,

$color(blue)(2^(nd)term=-3x$

$color(blue)(3^(rd)term=k$

Formula to find $3^(rd)term$ is :

$color(red)(3^(rd)term=(2^(nd)term)^2/(4 xx1^(st)term))...to(A)$

So,

$k=(-3x)^2/(4*x^2)=(9x^2)/(4x^2)=9/4$

Subst. $k=9/4$ into $(I)$ ,we get

$=>x^2-3x+9/4=9/4+18=81/4$

$=>(x-3/2)^2=(9/2)^2$

$=>x-3/2=+-9/2$

$=>x-3/2=9/2 orx-3/2=-9/2$

$=>x=9/2+3/2 or x=3/2-9/2$

$=>x=12/2 or x=-6/2$

$=>x=6 or x=-3$

......................................................................................

Note:
We can use Formula $(A)$ without any doubt.

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How do you solve $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ by completing the square?

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