Solving and Graphing Radical Equations - Math

Card 0 of 36

Question

Solve the equation for .

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

Answer

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

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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Question

Solve for :

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

Answer

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.

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

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Question

Solve the following radical expression:

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

Answer

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:

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Question

Solve the following radical expression:

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

Answer

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:

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Question

Solve for :

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

Answer

Begin by cubing both sides:

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

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

Now we can easily solve:

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Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the equation for .

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

Answer

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

Add to both sides.

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

Square both sides.

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

$\small 4x+9=169$

Isolate .

$\small 4x=160$

$\small x=40$

Compare your answer with the correct one above

Question

Solve for :

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

Answer

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.

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve for :

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

Answer

Begin by cubing both sides:

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

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

Now we can easily solve:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the following radical expression:

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

Answer

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:

Compare your answer with the correct one above

Question

Solve the equation for .

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

Answer

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

Add to both sides.

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

Square both sides.

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

$\small 4x+9=169$

Isolate .

$\small 4x=160$

$\small x=40$

Compare your answer with the correct one above

Question

Solve for :

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

Answer

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.

Compare your answer with the correct one above

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