Solving Radical Equations - Algebra 2

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Question

Solve for x: $\sqrt{2x+1} = 11$

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Answer

To solve, perform inverse opperations, keeping in mind order of opperations:

$\sqrt{2x+1} = 11$ first, square both sides

subtract 1

divide by 2

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Question

Solve for x: $(\sqrt{x} + 19 )^2 = 10,000$

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Answer

To solve, perform inverse opperations, keeping in mind order of opperations:

$(\sqrt x + 19 ) ^ 2 = 10,000$ take the square root of both sides

$\sqrt x + 19 = 100$ subtract 19 from both sides

$\sqrt{x } = 81$ square both sides

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Question

Solve for x: $5\sqrt{x - 12 } = 20$

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Answer

To solve, use inverse opperations keeping in mind order of opperations:

$5\sqrt {x-12} = 20$ divide both sides by 5

$\sqrt{x-12 } = 4$ square both sides

add 12 to both sides

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Question

Solve for :

$x - 2\sqrt{x} = 3$

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Answer

When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals. So a good first step is to add $2\sqrt{x}$ and subtract from both sides so that you get:

$x-3=2{\sqrt{x}}$

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

Then when you subtract from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

This factors to:

Which would seem to yield solutions of and . However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions. Here if you plug in the math holds: $9-2{\sqrt{9}}=3$ because $2\sqrt{9}=2(3)=6$ , and .

But if you plug in , you'll see that the original equation is not satisfied: $1-3\neq 2\sqrt{1}$ because does not equal . Therefore the only proper answer is .

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Question

Solve for :

$\sqrt{x-3} = x - 5$

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Answer

To solve, first square both sides:

$\sqrt{x-3}^2 = (x-5)^2$ squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply :

This is a quadratic, we just need to combine like terms and get it equal to 0

now we can solve using the quadratic formula:

$x = \frac{11 \pm \sqrt{11^2 - 4128}}{2 } = \frac{11 \pm \sqrt {121 - 112}}{2 } = \frac{11 \pm \sqrt{9 }}{2} = \frac{11 \pm 3 }{2}$

This gives us 2 potential answers:

$x = \frac{11 + 3 }{2 } = \frac{14 }{2} = 7$ and

$x = \frac{11 - 3 }{ 2 } = \frac{8}{2} = 4$

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Question

Solve for .

$\sqrt{x}=4$

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Answer

$\sqrt{x}=4$ To get rid of the radical, we square both sides.

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Question

Solve for .

$\sqrt{x}=-5$

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Answer

To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.

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Question

Solve for .

$\sqrt{x+1}=5$

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Answer

$\sqrt{x+1}=5$ Square both sides to get rid of the radical.

Subtract on both sides.

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Question

Solve the equation: $\sqrt{6-x}+2 = 16$

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Answer

Subtract two from both sides.

$\sqrt{6-x}+2 -(2)= 16-(2)$

Simplify the equation.

$\sqrt{6-x} = 14$

Square both sides.

$(\sqrt{6-x})^2 =( 14)^2$

Add on both sides.

Subtract on both sides.

The answer is:

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Question

Solve for .

$\sqrt{x-4}=7$

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Answer

$\sqrt{x-4}=7$ Square both sides to get rid of the radical.

Add on both sides.

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Question

Solve for .

$\sqrt{3x}=9$

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Answer

$\sqrt{3x}=9$ Square both sides to get rid of the radical.

Divide on both sides.

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Question

Solve for .

$\sqrt{\frac{x}{2}}=8$

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Answer

$\sqrt{{\frac{x}{2}}{}}=8$ Square both sides to get rid of the radical.

$\frac{x}{2}=8^2=64$ Multiply on both sides.

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Question

Solve: $\sqrt{3x-3}-3 = 3$

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Answer

Add three on both sides.

$\sqrt{3x-3}-3+3 = 3+3$

The equation becomes:

$\sqrt{3x-3} =6$

Square both sides to eliminate the radical.

$(\sqrt{3x-3} )^2=6^2$

Add three on both sides.

Divide by three on both sides.

$\frac{3x}{3}=\frac{39}{3}$

The answer is:

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Question

Solve the equation: $\sqrt{2x}+1 = \sqrt{x+3}$

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Answer

Subtract $\sqrt{2x}$ from both sides to group the radicals.

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

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

Square both sides.

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

Use the FOIL method to simplify the right side.

$(\sqrt{x+3})(\sqrt{x+3})+(\sqrt{x+3})(-\sqrt{2x})+(-\sqrt{2x})(\sqrt{x+3})+(-\sqrt{2x})(-\sqrt{2x})$

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

Combine like-terms.

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

Subtract one from both sides, and add $2\sqrt{2x(x+3)}$ on both sides.

The equation becomes: $2\sqrt{2x(x+3)}= 3x+2$

Divide by two on both sides and distribute the terms inside the radical.

$\sqrt{2x^2+6x} = \frac{3}{2}x+1$

Square both sides.

$(\sqrt{2x^2+6x} )^2=( \frac{3}{2}x+1)^2$

Simplify the right side by FOIL method.

$2x^2+6x = \frac{9}{4}x^2+3x+1$

Subtract on both sides. This is the same as subtracting $\frac{8}{4}x^2$ on both sides.

$2x^2+6x -2x^2 = (\frac{9}{4}x^2-\frac{8}{4}x^2)+3x+1$

$6x = \frac{1}{4}x^2+3x+1$

Subtract on both sides. The equation will become:

$0= \frac{1}{4}x^2-3x+1$

Multiply by four on both sides to eliminate the fractional denominator.

$(0)(4)= 4(\frac{1}{4}x^2-3x+1)$

Use the quadratic equation to solve for the roots.

$x = \frac{-(-12)\pm \sqrt{(-12)^2-4(1)(4)}}{2(1)}$

Simplify the radical and fraction.

$x = \frac{12\pm \sqrt{144-16}}{2} = \frac{12 \pm \sqrt{128}}{2} = \frac{12 \pm \sqrt{64} \cdot \sqrt{2}}{2}$

$x=\frac{12\pm 8\sqrt2}{2} = 6\pm 4\sqrt2$

Substitute the values of $6+4\sqrt{2}$ and $6-4\sqrt{2}$ back into the original equation, and only $6-4\sqrt{2}$ will satisfy both sides of the equation.

The answer is: $6-4\sqrt{2}$

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Question

Solve the equation: $2\sqrt{2x-3} = 4$

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Answer

Divide by two on both sides.

$\frac{2\sqrt{2x-3}}{2} = \frac{4}{2}$

The equation becomes: $\sqrt{2x-3} = 2$

Square both sides.

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

Add three on both sides.

Divide by two on both sides.

$\frac{2x}{2}=\frac{7}{2}$

The answer is: $\frac{7}{2}$

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Question

Solve for :

$x + 3 \sqrt{x} = 28$

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Answer

One way to solve this equation is to substitute for $\sqrt{x}$ and, subsequently, $u^{2}$ for :

$x + 3 \sqrt{x} = 28$

$u^{2} + 3 u = 28$

$u^{2} + 3 u - 28 = 0$

Solve the resulting quadratic equation by factoring the expression:

Set each linear binomial to sero and solve:

or

Substitute back:

$\sqrt{x}= -7$ - this is not possible.

$\sqrt{x}= 4$

- this is the only solution.

None of the responses state that is the only solution.

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Question

If $\sqrt{2x+3}-5 = -2$ , then ?

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Answer

We will begin by adding 5 to both sides of the equation.

$\\sqrt{2x+3}-5+5 = -2+5$

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

Next, let's square both sides to eliminate the radical.

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

Finally, we can solve this like a simple two-step equation. Subtract 3 from both sides of the equation.

Now, divide each side by 2.

$\frac{2x}{2}=\frac{6}{2}$

Finally, check the solution to make sure that it results in a true statement.

$\sqrt{2(3)+3}=3$

$\sqrt{9}=3$

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Question

Solve the following radical equation.

$\frac{10\sqrt{8}}{\sqrt{16}} + 3\sqrt{2}$

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Answer

We can simplify the fraction:

$\frac{10\sqrt{8}}{\sqrt{16}} = \frac{10\sqrt{4\times 2}}{4}=\frac{10(2\sqrt{2})}{4} =\frac{20\sqrt{2}}{4}=5\sqrt{2}$

Plugging this into the equation leaves us with:

$5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$

Note: Because they are like terms, we can add them.

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Question

Solve the following radical equation.

$\sqrt{x-6} =2$

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Answer

In order to solve this equation, we need to know that

$(\sqrt{x-6})^{2} = x-6$

How? Because of these two facts:

$\sqrt{x-6} = (x-6)^{\frac{1}{2}}$

Power rule of exponents: when we raise a power to a power, we need to mulitply the exponents:

$\frac{1}{2} \times2=1$

With this in mind, we can solve the equation:

$\sqrt{x-6} =2$

In order to eliminate the radical, we have to square it. What we do on one side, we must do on the other.

$(\sqrt{x-6})^{2} = 2^{2}$

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Question

Solve the following radical equation.

$\sqrt{\frac{x}{2}}=5$

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Answer

In order to solve this equation, we need to know that

$\bigg(\sqrt{\frac{x}{2}}\bigg)^{2} =\frac{x}{2}$

Note: This is due to the power rule of exponents.

With this in mind, we can solve the equation:

$\sqrt{\frac{x}{2}} =5$

In order to get rid of the radical we square it. Remember what we do on one side, we must do on the other.

$\bigg(\sqrt{\frac{x}{2}}\bigg)^{2} =5^{2}$

$\frac{x}{2} =25$

$\frac{2}{1} \bigg(\frac{x}{2}\bigg)=25\bigg(\frac{2}{1}\bigg)$

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