How do you solve $x^{2} - 4 {(\frac{3 - x}{2})}^{2} = - 33$ $x^ { 2} - 4( \frac { 3- x } { 2} ) ^ { 2} = - 33$ ?

Question

1200 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-29T16:26:46

Start by expanding the bracket completely, then simplify it and rearrange to solve.

The equation can be rewritten as this:
$x^{2} - 4 \frac{{(3 - x)}^{2}}{{(2)}^{2}} + 33 = 0$ $x^2-4(3-x)^2/(2)^2+33=0$

Expand the brackets
$x^{2} - 4 (\frac{9 - 6 x + x^{2}}{4}) + 33 = 0$ $x^2-4((9-6x+x^2)/4)+33=0$

The 4s cancel out, which gives you:
$x^{2} - (9 - 6 x + x^{2}) + 33 = 0$ $x^2-(9-6x+x^2)+33=0$

Expand and simplify:
$x^{2} - 9 + 6 x - x^{2} + 33 = 0$ $x^2-9+6x-x^2+33=0$
$24 + 6 x = 0$ $24+6x=0$

Solve:
$6 x = - 24$ $6x=-24$
$x = - \frac{24}{6}$ $x=-24/6$
$x = - 4$ $x=-4$

How do you solve x^ { 2} - 4( \frac { 3- x } { 2} ) ^ { 2} = - 33x2−4(3−x2)2=−33x^ { 2} - 4( \frac { 3- x } { 2} ) ^ { 2} = - 33?