Basic Squaring / Square Roots

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PSAT Math › Basic Squaring / Square Roots

Questions 1 - 10

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$-\sqrt{3}+9$

$\sqrt{63}$

$2\sqrt{3}$

$7\sqrt{3}+9$

Explanation

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{9*3}\rightarrow \sqrt{3^2*3}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{16*3}\rightarrow \sqrt{4^2*3}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$-\sqrt{3}+9$

$\sqrt{63}$

$2\sqrt{3}$

$7\sqrt{3}+9$

Explanation

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{9*3}\rightarrow \sqrt{3^2*3}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{16*3}\rightarrow \sqrt{4^2*3}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$-\sqrt{3}+9$

$\sqrt{63}$

$2\sqrt{3}$

$7\sqrt{3}+9$

Explanation

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{9*3}\rightarrow \sqrt{3^2*3}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{16*3}\rightarrow \sqrt{4^2*3}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify. Assume all integers are positive real numbers.

$\sqrt[3]{24x^{16}y^{12}}$

$2x^{5}y^{4}\sqrt[3]{3x}$

$8x^{16}y^{12}$

$2x^{8}y^{4}\sqrt[3]{3x^{2}}$

$2x^{2}y^{3}\sqrt[3]{3xy}$

$2x^{5}y^{4}$

Explanation

$\sqrt[3]{24x^{16}y^{12}}$

Index of means the cube root of Radican $24x^{16}y^{12}$

Find a perfect cube in $\rightarrow$ $\sqrt[3]{8\cdot 3}$

Simplify the perfect cube, giving you $2\sqrt[3]{3}$ .

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$

$y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$

The final answer would be

$2x^{5}y^{4}\sqrt[3]{3x}$

Simplify. Assume all integers are positive real numbers.

$\sqrt[3]{24x^{16}y^{12}}$

$2x^{5}y^{4}\sqrt[3]{3x}$

$8x^{16}y^{12}$

$2x^{8}y^{4}\sqrt[3]{3x^{2}}$

$2x^{2}y^{3}\sqrt[3]{3xy}$

$2x^{5}y^{4}$

Explanation

$\sqrt[3]{24x^{16}y^{12}}$

Index of means the cube root of Radican $24x^{16}y^{12}$

Find a perfect cube in $\rightarrow$ $\sqrt[3]{8\cdot 3}$

Simplify the perfect cube, giving you $2\sqrt[3]{3}$ .

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$

$y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$

The final answer would be

$2x^{5}y^{4}\sqrt[3]{3x}$

Simplify. Assume all integers are positive real numbers.

$\sqrt[3]{24x^{16}y^{12}}$

$2x^{5}y^{4}\sqrt[3]{3x}$

$8x^{16}y^{12}$

$2x^{8}y^{4}\sqrt[3]{3x^{2}}$

$2x^{2}y^{3}\sqrt[3]{3xy}$

$2x^{5}y^{4}$

Explanation

$\sqrt[3]{24x^{16}y^{12}}$

Index of means the cube root of Radican $24x^{16}y^{12}$

Find a perfect cube in $\rightarrow$ $\sqrt[3]{8\cdot 3}$

Simplify the perfect cube, giving you $2\sqrt[3]{3}$ .

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$

$y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$

The final answer would be

$2x^{5}y^{4}\sqrt[3]{3x}$

If $\sqrt{x}=3^2$ what is $x$ ?

$81$

$9$

$3$

$729$

$27$

Explanation

Square both sides:

x = (32)2 = 92 = 81

If $\sqrt{x}=3^2$ what is $x$ ?

$81$

$9$

$3$

$729$

$27$

Explanation

Square both sides:

x = (32)2 = 92 = 81

If $\sqrt{x}=3^2$ what is $x$ ?

$81$

$9$

$3$

$729$

$27$

Explanation

Square both sides:

x = (32)2 = 92 = 81

The square root(s) of 36 is/are ________.

6 and -6

-6

6, -6, and 0

None of these answers are correct.

Explanation

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

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