Squaring / Square Roots / Radicals

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PSAT Math › Squaring / Square Roots / Radicals

Questions 1 - 10

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

Hero

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?

12√5

4√14

8√14

14√2

48√77

Explanation

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.

Hero

Hero2

Simplify the radical expression.

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

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

Explanation

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

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

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

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

Simplify.

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

Simplify the radical expression.

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

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

Explanation

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

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

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

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

Simplify.

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

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

Hero

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?

12√5

4√14

8√14

14√2

48√77

Explanation

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.

Hero

Hero2

Simplify the radical.

$\sqrt{3283}$

$7\sqrt{67}$

$56$

$57.3$

$7\sqrt{63}$

$67\sqrt{49}$

Explanation

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

Simplify the radical.

$\sqrt{3283}$

$7\sqrt{67}$

$56$

$57.3$

$7\sqrt{63}$

$67\sqrt{49}$

Explanation

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

Expand:

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

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

$100x^{2} + 2x + \frac{1}{10}$

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

$100x^{2} + \frac{1}{10}$

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

Explanation

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

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

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

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

Expand:

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

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

$100x^{2} + 2x + \frac{1}{10}$

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

$100x^{2} + \frac{1}{10}$

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

Explanation

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

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

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

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

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

There is no $x^{3}$ term in the expansion of $\left (5 x^{2 } - x + 1\right )^{2}$ .

Explanation

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

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

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

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

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

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

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

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

Expand:

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

$100x^{2} + 16x + 0.64$

$100x^{2} + 1.6x + 6.4$

$100x^{2} +0. 16x + 0.064$

$100x^{2}+8x + 0.64$

$100x^{2} +0.8x+ 6.4$

Explanation

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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