Math - Radicals | Practice Hub

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.

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

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

Convert the radical to exponential notation.

$\small \sqrt[4]{13}$

$13^{\frac{1}{4}}$

$\small \sqrt{13^4}$

$\small 169$

$\small 13^4$

Explanation

Remember that any term outside the radical will be in the denominator of the exponent.

$\sqrt[b]{x^a}=x^{\frac{a}{b}}$

Since does not have any roots, we are simply raising it to the one-fourth power.

$\sqrt[4]{13}=13^{\frac{1}{4}}$

Convert the radical to exponential notation.

$\small \sqrt[4]{13}$

$13^{\frac{1}{4}}$

$\small \sqrt{13^4}$

$\small 169$

$\small 13^4$

Explanation

Remember that any term outside the radical will be in the denominator of the exponent.

$\sqrt[b]{x^a}=x^{\frac{a}{b}}$

Since does not have any roots, we are simply raising it to the one-fourth power.

$\sqrt[4]{13}=13^{\frac{1}{4}}$

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

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

Convert the radical to exponential notation.

$\small \sqrt[4]{13}$

$13^{\frac{1}{4}}$

$\small \sqrt{13^4}$

$\small 169$

$\small 13^4$

Explanation

Remember that any term outside the radical will be in the denominator of the exponent.

$\sqrt[b]{x^a}=x^{\frac{a}{b}}$

Since does not have any roots, we are simply raising it to the one-fourth power.

$\sqrt[4]{13}=13^{\frac{1}{4}}$

Factor and simplify the following radical expression:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$3abc\sqrt[4]{6ab^3c^2}$

$3a^2bc\sqrt[4]{6ab^3c^2}$

$3ab^2c\sqrt[4]{6ab^3c^2}$

$3abc^2\sqrt[4]{6ab^3c^2}$

$3abc\sqrt[4]{6a^2b^3c^2}$

Explanation

Begin by converting the radical into exponent form:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$(18a^3bc^5)^{\frac{1}{4}}\cdot ({27a^2b^6c})^{\frac{1}{4}}$

Now, combine the bases:

$(486a^5b^7c^6)^{\frac{1}{4}}$

Simplify the integer:

$(2\cdot 3\cdot 3^4a^5b^7c^6)^{\frac{1}{4}}$

Now, simplify the exponents:

$(2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4)^{\frac{1}{4}}$

Convert back into radical form and simplify:

$\sqrt[4]{2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4}$

$3abc\sqrt[4]{2\cdot 3ab^3c^2}$

$3abc\sqrt[4]{6ab^3c^2}$

Factor and simplify the following radical expression:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$3abc\sqrt[4]{6ab^3c^2}$

$3a^2bc\sqrt[4]{6ab^3c^2}$

$3ab^2c\sqrt[4]{6ab^3c^2}$

$3abc^2\sqrt[4]{6ab^3c^2}$

$3abc\sqrt[4]{6a^2b^3c^2}$

Explanation

Begin by converting the radical into exponent form:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$(18a^3bc^5)^{\frac{1}{4}}\cdot ({27a^2b^6c})^{\frac{1}{4}}$

Now, combine the bases:

$(486a^5b^7c^6)^{\frac{1}{4}}$

Simplify the integer:

$(2\cdot 3\cdot 3^4a^5b^7c^6)^{\frac{1}{4}}$

Now, simplify the exponents:

$(2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4)^{\frac{1}{4}}$

Convert back into radical form and simplify:

$\sqrt[4]{2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4}$

$3abc\sqrt[4]{2\cdot 3ab^3c^2}$

$3abc\sqrt[4]{6ab^3c^2}$

Factor and simplify the following radical expression:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$3abc\sqrt[4]{6ab^3c^2}$

$3a^2bc\sqrt[4]{6ab^3c^2}$

$3ab^2c\sqrt[4]{6ab^3c^2}$

$3abc^2\sqrt[4]{6ab^3c^2}$

$3abc\sqrt[4]{6a^2b^3c^2}$

Explanation

Begin by converting the radical into exponent form:

$\sqrt[4]{18a^3bc^5}\cdot \sqrt[4]{27a^2b^6c}$

$(18a^3bc^5)^{\frac{1}{4}}\cdot ({27a^2b^6c})^{\frac{1}{4}}$

Now, combine the bases:

$(486a^5b^7c^6)^{\frac{1}{4}}$

Simplify the integer:

$(2\cdot 3\cdot 3^4a^5b^7c^6)^{\frac{1}{4}}$

Now, simplify the exponents:

$(2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4)^{\frac{1}{4}}$

Convert back into radical form and simplify:

$\sqrt[4]{2\cdot 3\cdot 3^4aa^4b^3b^4c^2c^4}$

$3abc\sqrt[4]{2\cdot 3ab^3c^2}$

$3abc\sqrt[4]{6ab^3c^2}$

Factor and simplify the following radical expression:

$(4-2\sqrt{2})(\sqrt{2}-4)$

$12\sqrt{2}-20$

$10\sqrt{2}-20$

$12\sqrt{2}-18$

$12\sqrt{2}-16$

$14\sqrt{2}-20$

Explanation

Begin by using the FOIL method (First Outer Inner Last) to expand the expression.

$(4-2\sqrt{2})(\sqrt{2}-4)$

$4\sqrt{2}-16-4+8\sqrt{2}$

Now, combine like terms:

$12\sqrt{2}-20$

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