Radical equations

Example 1: Solve $\sqrt{3x-8}=\sqrt{7x-11}$

Solution:

There is a square root on both the left-hand side and the right-hand side, so we can remove the square roots by squaring both sides,

$\begin{align*} (\sqrt{3x-8})^2&=(\sqrt{7x-11})^2\\ 3x-8 &=7x-11\\ -8+11 &=7x-3x\\ 3 &=4x\\ x &=\dfrac{3}{4}\\ \end{align*}$

Example 2: Solve $\sqrt{4u-3}+2=u$

Solution:

Keep the radical on one side and move everything else to the other side,

$\sqrt{4u-3}=u-2$

Square both sides, which removes the radical sign:

$4u-3=(u-2)^2$

$4u-3=u^2-4u+4$

$0=u^2-8u+7$

or

$u^2-8u+7=0$

Factoring,

$(u-7)(u-1)=0$

or

$u=7$ or $u=1$

There are two roots; substitute these values into the original equation and check.

Substituting $u=7$ into the original equation,

$\sqrt{4\cdot 7-3}+2=7$

$\sqrt{25}+2=7$

$5+2=7$

$7=7$

This statement is true; therefore, $7$ is a solution of the given equation.

Next, substitute the other value, $u=1$.

$\sqrt{4\cdot 1-3}+2=1$

$\sqrt{4-3}+2=1$

$\sqrt{1}+2=1$

$3=1$

This statement is not true. Therefore, $u=1$ is not a solution of the given equation.

So, the solution of the equation is $u=7$.

Example 3: Solve $\sqrt[3]{5y-9}\:-5=-3$

Solution:

$\sqrt[3]{5y-9}\:-5=-3$

Keep the radical on one side and move everything to the other side,

$\sqrt[3]{5y-9}=5-3$

$\sqrt[3]{5y-9}=2$

Raise both sides to the power $3$,

$(\sqrt[3]{5y-9})^3=2^3$

$5y-9=8$

Solving for $y$,

$y=\dfrac{17}{5}$

Example 4: Solve $(13y-10)^{1/4}+7=2$

Solution:

$(13y-10)^{1/4}+7=2$

Keep the rational exponent on one side and move everything to the other side,

$(13y-10)^{1/4}=2-7$

$(13y-10)^{1/4}=-7$

A fourth root is on the left-hand side and a negative number on the right-hand side. Since a fourth root cannot be negative, the equation has no solution.

Example 5: Solve $(3w-10)^3=125$

Solution:

$(3w-10)^3=125$

Taking the cube root of both sides,

$3w-10=5$

Solving for $w$,

$\boxed{w=5}$