How do you solve the equation $2 (x^{2} - 5) = - x^{2} - 1$ $2(x^2-5)=-x^2-1$ ?

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Answer 1 · 2017-06-08T21:14:25

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

Multiply the $2$ $2$ through on the left hand side:

$2 (x^{2} - 5) = - x^{2} - 1$ $2(x^2-5)=-x^2-1$

$2 x^{2} - 10 = - x^{2} - 1$ $2x^2-10=-x^2-1$

Add $x^{2}$ $color(red)(x^2)$ and $10$ $color(blue)(10)$ to both sides

$2 x^{2} + x^{2} - 10 + 10 = - x^{2} + x^{2} - 1 + 10$ $2x^2color(red)(+x^2)-10color(blue)(+10)=-x^2color(red)(+x^2)-1color(blue)(+10)$

$3 x^{2} = 9$ $3x^2=9$

Divide both sides by $3$ $3$

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

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

Take $\sqrt{a a}$ $sqrt(color(white)(aa))$ of both sides

$\sqrt{x^{2}} = \pm \sqrt{3}$ $color(red)(sqrt(color(black)(x^2))=color(red)(+-sqrt(color(black)(3))$

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

How do you solve the equation 2(x^2-5)=-x^2-12(x2−5)=−x2−12(x^2-5)=-x^2-1?