Solving Radical Equations and Inequalities

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Math › Solving Radical Equations and Inequalities

Questions 1 - 9

1

Solve for :

$\sqrt{x^2+12x+36} =4$

$x = \left { -10, -2\right }$

Explanation

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left { -10, -2\right }$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

2

Solve the following radical expression:

$\sqrt{y+12}+1=\sqrt{y+21}$

Explanation

Begin by squaring both sides of the equation:

$\sqrt{y+12}+1=\sqrt{y+21}$

$(\sqrt{y+12}+1)^2=(\sqrt{y+21})^2$

$(y+12)+2\sqrt{y+12}+1=y+21$

Now, combine like terms and simplify:

$2\sqrt{y+12}=8$

$\sqrt{y+12}=4$

Once again, take the square of both sides of the equation:

$(\sqrt{y+12}=4)^2$

Solve the linear equation:

3

Solve the following radical expression:

$\sqrt{4x+29}=(x+6)$

No real solutions

Explanation

Begin by squaring both sides of the equation:

$\sqrt{4x+29}=(x+6)$

$(\sqrt{4x+29})^2=(x+6)^2$

Now, combine like terms:

Factor the equation:

However, when plugging in the values, does not work. Therefore, there is only one solution:

4

Solve the following radical expression:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

Explanation

Begin by squaring both sides of the equation:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

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

$(2x+1)-4\sqrt{2x+1}+4=2x-7$

Combine like terms:

$-4\sqrt{2x+1}=-12$

$\sqrt{2x+1}=3$

Once again, square both sides of the equation:

$(\sqrt{2x+1}=3)^2$

Solve the linear equation:

5

Solve for :

$\sqrt[3]{y+1}=3$

Explanation

Begin by cubing both sides:

$\sqrt[3]{y+1}=3$

$(\sqrt[3]{y+1}=3)^3$

Now we can easily solve:

6

Solve the following radical expression:

$\sqrt{5+2x}=x-5$

Explanation

Begin by taking the square of both sides:

$\sqrt{5+2x}=x-5$

$(\sqrt{5+2x})^2=(x-5)^2$

Combine like terms:

Factor the equation and solve:

However, when plugging in the values, does not work. Therefore, there is only one solution:

7

Solve the following radical expression:

$\sqrt{x+11}-x=-9$

Explanation

To solve the radical expression, begin by subtracting from each side of the equation:

$\sqrt{x+11}-x=-9$

$\sqrt{x+11}=x-9$

Now, square both sides of the equation:

$(\sqrt{x+11})^2=(x-9)^2$

Combine like terms:

Factor the expression and solve:

However, when plugged into the original equation, does not work because the radical cannot be negative. Therefore, there is only one solution:

8

Solve the following radical expression:

$\sqrt{3x+1}+2=7$

Explanation

Begin by subtracting from each side of the equation:

$\sqrt{3x+1}+2=7$

$\sqrt{3x+1}=5$

Now, square the equation:

$(\sqrt{3x+1}=5)^2$

Solve the linear equation:

9

Solve the equation for .

$\sqrt{4x+9}-8=5$

Explanation

$\small \sqrt{4x+9}-8=5$

Add to both sides.

$\small \sqrt{4x+9}=13$

Square both sides.

$\small (\sqrt{4x+9})^2=13^2$

$\small 4x+9=169$

Isolate .

$\small 4x=160$

$\small x=40$

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