How do you solve $- 4 x^{2} + x^{2} = - 36$ $-4x^2 + x^2 = -36$ ?

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Answer 1 · 2016-07-04T21:27:44

$x = \pm 2 \sqrt{3}$ $x=+-2sqrt(3)$

given: $- 4 x^{2} + x^{2} = - 36$ $" "color(green)(-4x^2)color(brown)(+x^2)=-36$

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One way of calculating this

We know that $+ 3 + 1 = + 4$ $" "+3 + 1 = +4$

So $- 3 - 1 = - 4$ $" -3-1 = -4$

Write $- 4 x^{2}$ $-4x^2$ as $- 3 x^{2} - 1 x^{2}$ $-3x^2-1x^2$

Mathematically should be written as $- 3 x^{2} - x^{2}$ $color(green)(-3x^2-x^2)$
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So we know have:

$- 3 x^{2} - x^{2} + x^{2} = - 36$ $color(green)(-3x^2-x^2)color(brown)(+x^2)=-36$

But $- x^{2} + x^{2} = 0$ $-x^2+x^2=0$ giving

$- 3 x^{2} + 0 = - 36$ $-3x^2+0=-36$

$- 3 x^{2} = - 36$ $-3x^2=-36$

Divide both sides by $- 3$ $-3$

$\frac{- 3}{- 3} x^{2} = \frac{- 36}{- 3}$ $(-3)/(-3)x^2=(-36)/(-3)$

Divide one negative number by another negative number and you have a positive answer.

But $\frac{- 3}{- 3} = + 1 and \frac{- 36}{- 3} = + 12$ $(-3)/(-3)=+1" and "(-36)/(-3)= + 12$

$1 \times x^{2} = 12$ $1xx x^2=12$

$x^{2} = 12$ $x^2=12$

Square root both sides

$\sqrt{x^{2}} = \sqrt{3 \times 2^{2}}$ $sqrt(x^2)=sqrt(3xx2^2)$

$x = \pm 2 \sqrt{3}$ $x=+-2sqrt(3)$

How do you solve −4x2+x2=−36-4x^2 + x^2 = -36?