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# Equations with Square Roots Calculator

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###  Difficult Problems

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:

$\sqrt{x}+\sqrt{x+7}=7$
2

We need to isolate the dependent variable , we can do that by simultaneously subtracting $\sqrt{x+7}$ from both sides of the equation

$\sqrt{x}=7-\sqrt{x+7}$

Removing the variable's exponent raising both sides of the equation to the power of $2$

$\left(\sqrt{x}\right)^{\frac{1}{\frac{1}{2}}}=\left(7-\sqrt{x+7}\right)^{\frac{1}{\frac{1}{2}}}$

Divide $1$ by $\frac{1}{2}$

$\left(\sqrt{x}\right)^{2}=\left(7-\sqrt{x+7}\right)^{\frac{1}{\frac{1}{2}}}$

Simplify $\left(\sqrt{x}\right)^{2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $\frac{1}{2}$ and $n$ equals $2$

$x^{\frac{1}{2}\cdot 2}$

Multiply $\frac{1}{2}$ times $2$

$x$

Multiply $\frac{1}{2}$ times $2$

$x=\left(7-\sqrt{x+7}\right)^{\frac{1}{\frac{1}{2}}}$

Divide $1$ by $\frac{1}{2}$

$x=\left(7-\sqrt{x+7}\right)^{2}$
3

Removing the variable's exponent raising both sides of the equation to the power of $2$

$x=\left(7-\sqrt{x+7}\right)^{2}$

Square of the first term: $\left(7\right)^2 = . Double product of the first by the second:$2\left(7\right)\left(-\sqrt{x+7}\right) = .

Square of the second term: $\left(-\sqrt{x+7}\right)^2 = Expand$\left(7-\sqrt{x+7}\right)^{2}7^2+2\cdot 7\left(-1\right)\sqrt{x+7}+\left(-\sqrt{x+7}\right)^2$Simplify$\left(-\sqrt{x+7}\right)^27^2+2\cdot 7\left(-1\right)\sqrt{x+7}+\left(\sqrt{x+7}\right)^2$Simplify$\left(\sqrt{x+7}\right)^2$using the power of a power property:$\left(a^m\right)^n=a^{m\cdot n}$. In the expression,$m$equals$\frac{1}{2}$and$n$equals$27^2+2\cdot 7\left(-1\right)\sqrt{x+7}+x+7$4 Expand$\left(7-\sqrt{x+7}\right)^{2}x=49-14\sqrt{x+7}+x+7$5 Add the values$49$and$7x=56-14\sqrt{x+7}+x$6 Move the term with the square root to the left side of the equation, and all other terms to the right side. Remember to change the signs of each term$14\sqrt{x+7}=56+x-x$7 Cancel like terms$x$and$-x14\sqrt{x+7}=56$Removing the variable's exponent raising both sides of the equation to the power of$2\left(14\sqrt{x+7}\right)^{\frac{1}{\frac{1}{2}}}=56^{\frac{1}{\frac{1}{2}}}$Divide$1$by$\frac{1}{2}\left(14\sqrt{x+7}\right)^{2}=56^{\frac{1}{\frac{1}{2}}}$Divide$1$by$\frac{1}{2}\left(14\sqrt{x+7}\right)^{2}=56^{2}$Calculate the power$56^{2}\left(14\sqrt{x+7}\right)^{2}=3136$8 Removing the variable's exponent raising both sides of the equation to the power of$2\left(14\sqrt{x+7}\right)^{2}=3136$The power of a product is equal to the product of it's factors raised to the same power$14^{2}\left(\sqrt{x+7}\right)^{2}=3136$Calculate the power$14^{2}196\left(\sqrt{x+7}\right)^{2}=3136$Simplify$\left(\sqrt{x}\right)^{2}$using the power of a power property:$\left(a^m\right)^n=a^{m\cdot n}$. In the expression,$m$equals$\frac{1}{2}$and$n$equals$2x^{\frac{1}{2}\cdot 2}$Multiply$\frac{1}{2}$times$2x$Simplify$\left(\sqrt{x+7}\right)^2$using the power of a power property:$\left(a^m\right)^n=a^{m\cdot n}$. In the expression,$m$equals$\frac{1}{2}$and$n$equals$2\left(x+7\right)^{\frac{1}{2}\cdot 2}$Multiply$\frac{1}{2}$times$2x+7$Simplify$\left(\sqrt{x+7}\right)^{2}$using the power of a power property:$\left(a^m\right)^n=a^{m\cdot n}$. In the expression,$m$equals$\frac{1}{2}$and$n$equals$2196\left(x+7\right)^{\frac{1}{2}\cdot 2}$Multiply$\frac{1}{2}$times$2196\left(x+7\right)$Multiply$\frac{1}{2}$times$2196\left(x+7\right)=3136$9 The power of a product is equal to the product of it's factors raised to the same power$196\left(x+7\right)=3136$10 Divide both sides of the equation by$196\frac{196\left(x+7\right)}{196}=\frac{3136}{196}$11 Simplifying the quotients$x+7=\frac{3136}{196}$12 Divide$3136$by$196x+7=16$We need to isolate the dependent variable , we can do that by simultaneously subtracting$7$from both sides of the equation$x=16-7$Subtract the values$16$and$-7x=9$13 We need to isolate the dependent variable , we can do that by simultaneously subtracting$7$from both sides of the equation$x=9$Verify that the solutions obtained are valid in the initial equation 14 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$x=9$##  Final answer to the problem$x=9\$

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