Squaring / Square Roots / Radicals - PSAT Math

Card 0 of 63

Question

x2 = 36

Quantity A: x

Quantity B: 6

Answer

x2 = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

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Question

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

Answer

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

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Question

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Answer

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}2\sqrt{3})+(\sqrt{5}\sqrt{12})$

Combine terms under radicals.

$2\sqrt{35}+\sqrt{125}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

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Question

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

Answer

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}\sqrt[3]{x^6}\sqrt[3]{y^{12}}$

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

Simplify.

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

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Question

Simplify the radical.

$\sqrt{3283}$

Answer

We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

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Question

Expand:

$\left ( 10x+ \frac{1}{5}\right )^{2}$

Answer

Use the perfect square trinomial pattern, setting $A = 10x , B = \frac{1}{5}$ :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$\left ( 10x + \frac{1}{5} \right )^{2} = (10x)^{2} + 2 (10x) \left ( \frac{1}{5} \right ) + \left ( \frac{1}{5} \right ) ^{2}$

$= 100x^{2} + 4x + \frac{1}{25}$

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Question

Expand:

$\left ( 10x+ 0.8\right )^{2}$

Answer

Use the perfect square trinomial pattern, setting :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$(10x + 0.8)^{2} = \left (10x \right )^{2} + 2 \cdot 10x \cdot 0.8 + 0.8^{2}$

$(10x + 0.8)^{2} = 100x^{2} + 16x + 0.64$

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Question

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{2}$ ?

Answer

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{2}$ is therefore 11.

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Question

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{3}$ ?

Answer

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{3}$ is therefore 10.

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Question

x2 = 36

Quantity A: x

Quantity B: 6

Answer

x2 = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

Compare your answer with the correct one above

Question

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

Answer

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

Compare your answer with the correct one above

Question

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Answer

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}2\sqrt{3})+(\sqrt{5}\sqrt{12})$

Combine terms under radicals.

$2\sqrt{35}+\sqrt{125}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

Compare your answer with the correct one above

Question

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

Answer

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}\sqrt[3]{x^6}\sqrt[3]{y^{12}}$

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

Simplify.

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

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Question

Simplify the radical.

$\sqrt{3283}$

Answer

We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Compare your answer with the correct one above

Question

Expand:

$\left ( 10x+ \frac{1}{5}\right )^{2}$

Answer

Use the perfect square trinomial pattern, setting $A = 10x , B = \frac{1}{5}$ :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$\left ( 10x + \frac{1}{5} \right )^{2} = (10x)^{2} + 2 (10x) \left ( \frac{1}{5} \right ) + \left ( \frac{1}{5} \right ) ^{2}$

$= 100x^{2} + 4x + \frac{1}{25}$

Compare your answer with the correct one above

Question

Expand:

$\left ( 10x+ 0.8\right )^{2}$

Answer

Use the perfect square trinomial pattern, setting :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$(10x + 0.8)^{2} = \left (10x \right )^{2} + 2 \cdot 10x \cdot 0.8 + 0.8^{2}$

$(10x + 0.8)^{2} = 100x^{2} + 16x + 0.64$

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Question

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{2}$ ?

Answer

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{2}$ is therefore 11.

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Question

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{3}$ ?

Answer

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{3}$ is therefore 10.

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Question

Simplify the radical.

$\sqrt{3283}$

Answer

We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Compare your answer with the correct one above

Question

x2 = 36

Quantity A: x

Quantity B: 6

Answer

x2 = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

Compare your answer with the correct one above

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