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

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Answer 1 · 2017-06-08T21:14:25

$x=+-sqrt(3)$

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

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

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

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

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

$3x^2=9$

Divide both sides by $3$

$(3x^2)/color(red)(3)=9/color(red)(3)$

$x^2=3$

Take $sqrt(color(white)(aa))$ of both sides

$color(red)(sqrt(color(black)(x^2))=color(red)(+-sqrt(color(black)(3))$

$x=+-sqrt(3)$

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