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Equations with Square Roots Calculator

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1

Here, we show you a step-by-step solved example of equations with square roots. This solution was automatically generated by our smart calculator:

$sqrt\left(x^2\:-\:1\right)\:=\:sqrt\left(x\:+\:2\right)$

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{0.5}$

$\left(x^2-1\right)^{0.5}^{\frac{1}{0.5}}=\left(x+2\right)^{0.5}^{\frac{1}{0.5}}$

Cancel exponents $0.5$ and $1$

$x^2-1=\left(x+2\right)^{0.5}^{\frac{1}{0.5}}$

Cancel exponents $0.5$ and $1$

$x^2-1=x+2$
2

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{0.5}$

$x^2-1=x+2$
3

Group the terms of the equation by moving the terms that have the variable $x$ to the left side, and those that do not have it to the right side

$x^2-x=2+1$
4

Add the values $2$ and $1$

$x^2-x=3$
5

Rewrite the equation

$x^2-x-3=0$
6

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where in this case $a=1$, $b=-1$ and $c=-3$. Then substitute the values of the coefficients of the equation in the quadratic formula: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{- -1\pm \sqrt{{\left(-1\right)}^2-4\cdot -3}}{2}$

Simplifying

$x=\frac{- -1\pm \sqrt{{\left(-1\right)}^2-4\cdot -3}}{2}$

Multiply $-1$ times $-1$

$x=\frac{1\pm \sqrt{{\left(-1\right)}^2-4\cdot -3}}{2}$

Multiply $-4$ times $-3$

$x=\frac{1\pm \sqrt{{\left(-1\right)}^2+12}}{2}$

Calculate the power ${\left(-1\right)}^2$

$x=\frac{1\pm \sqrt{1+12}}{2}$

Add the values $1$ and $12$

$x=\frac{1\pm \sqrt{13}}{2}$
7

Simplifying

$x=\frac{1\pm \sqrt{13}}{2}$
8

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x=\frac{1+\sqrt{13}}{2},\:x=\frac{1-\sqrt{13}}{2}$
9

Combining all solutions, the $2$ solutions of the equation are

$x=\frac{1+\sqrt{13}}{2},\:x=\frac{1-\sqrt{13}}{2}$

Verify that the solutions obtained are valid in the initial equation

10

The valid solutions to the equation are the ones that, when replaced in the original equation, don't result in any square root of a negative number and make both sides of the equation equal to each other

$x=\frac{1+\sqrt{13}}{2},\:x=\frac{1-\sqrt{13}}{2}$

Final answer to the exercise

$x=\frac{1+\sqrt{13}}{2},\:x=\frac{1-\sqrt{13}}{2}$

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