Squaring / Square Roots / Radicals - PSAT Math

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Question

x2 = 36

Quantity A: x

Quantity B: 6

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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?

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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}$)$

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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.

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Answer

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

Simplify.

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Question

Simplify the radical.

$\sqrt{3283}$

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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}$$

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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}$$

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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 ?

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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 is therefore 11.

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Question

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

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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 is therefore 10.

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Question

Simplify the radical.

$\sqrt{3283}$

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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}$$

Tap to reveal answer

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}$$

Tap to reveal answer

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 ?

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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 is therefore 11.

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Question

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

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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 is therefore 10.

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Question

Simplify the radical.

$\sqrt{3283}$

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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}$$

Tap to reveal answer

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}$$

Tap to reveal answer

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 ?

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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 is therefore 11.

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Question

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

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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 is therefore 10.

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Question

Simplify the radical.

$\sqrt{3283}$

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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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