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Algebra II : Intermediate Single-Variable Algebra

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Example Questions

Example Question #331 : Quadratic Equations And Inequalities

Use the quadratic formula to determine a root: y=3x^2-x+3

Possible Answers:

$\frac{1-i\sqrt{37}}{2}$

$\frac{1-i\sqrt{37}}{6}$

$\frac{1-i\sqrt{35}}{2}$

$\frac{1-i\sqrt{37}}{3}$

$\frac{1-i\sqrt{35}}{6}$

Correct answer:

$\frac{1-i\sqrt{35}}{6}$

Explanation:

Write the quadratic formula for polynomials in the form of y=ax^2+bx+c

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the known values.

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(3)}}{2(3)}$

Simplify the equation.

$x=\frac{1\pm\sqrt{1-36}}{6} = \frac{1\pm\sqrt{-35}}{6}= \frac{1\pm i\sqrt{35}}{6}$

The roots to this parabola is: $x= \frac{1\pm i\sqrt{35}}{6}$

The answer is $\frac{1-i\sqrt{35}}{6}$ .

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Example Question #501 : Intermediate Single Variable Algebra

Find a root using the quadratic equation: y=3-10x^2

Possible Answers:

$\frac{\sqrt{15}}{3}$

$- \frac{\sqrt{30}}{3}$

$- \frac{\sqrt{30}}{10}$

$- \frac{\sqrt{30}}{2}$

$- \frac{2\sqrt{15}}{3}$

Correct answer:

$- \frac{\sqrt{30}}{10}$

Explanation:

Write the quadratic equation.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Rewrite the given equation in y=ax^2+bx+c form.

y=3-10x^2 = -10x^2+0x+3

We can determine the coefficients of the terms.

a=-10

b=0

c=3

Substitute these values into the quadratic equation.

$x=\frac{\pm \sqrt{-4(-10)(3)}}{2(-10)} = \frac{\pm \sqrt{120}}{-20}$

Rewrite this fraction using common factors, and simplify each step.

$\frac{\pm \sqrt{120}}{-20} = \frac{\pm \sqrt{4\times 30}}{-20} = \frac{\pm \sqrt{4}\cdot\sqrt{30}}{-20}=\frac{\pm 2\cdot\sqrt{30}}{-20} =\pm \frac{\sqrt{30}}{10}$

One of the possible answers given is: $- \frac{\sqrt{30}}{10}$

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Example Question #44 : Quadratic Formula

Solve the following equation:

x^4-13x^2+36=0

Possible Answers:

$x=\pm3$

x=-2,2

$x=3,\pm2$

$x=\pm3,\pm2$

$x=\pm3,2$

Correct answer:

$x=\pm3,\pm2$

Explanation:

Let y=x^2 . Then the given equation can be rewritten in terms of as follows:

(x^2)^2-13(x^2)+36=0

$\Rightarrow y^2-13y+36=0$ .

By the quadratic formula,

$y=\frac{13\pm\sqrt{(13^2)-4(1)(36)}}{2(1)}$

$=\frac{13\pm5}{2}$

=9,4

Since y=x^2 , we now have x^2=9 and x^2=4 , which implies that $x=\pm3$ and $x=\pm2$ . Substituting each of these values yields a true statement; hence, the solutions to the original equation are

$x=\pm3,\pm2$ .

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Example Question #41 : Quadratic Formula

Solve for the roots: y=8x^2-4x-2

Possible Answers:

$\frac{-1\pm\sqrt{5}}{4}$

$\frac{-1\pm i\sqrt{5}}{4}$

$\frac{1\pm\sqrt{5}}{4}$

$\frac{1\pm\sqrt{5}}{16}$

$\frac{-1\pm\sqrt{5}}{2}$

Correct answer:

$\frac{1\pm\sqrt{5}}{4}$

Explanation:

Write the quadratic formula.

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The given equation y=8x^2-4x-2 is in standard form of:

a=ax^2+bx+c

The coefficients correspond to the values that go inside the quadratic equation.

a=8, b=-4, c=-2

Substitute the values into the equation.

$x= \frac{-(-4)\pm\sqrt{(-4)^2-4(8)(-2)}}{2(8)}$

Simplify this equation.

$x= \frac{4\pm\sqrt{16+64}}{16} = \frac{4\pm\sqrt{80}}{16}$

The radical can be rewritten as:

$\sqrt{80} = \sqrt{16}\cdot \sqrt 5 = 4\sqrt5$

Substitute and simplify the fraction.

$x=\frac{4\pm4\sqrt{5}}{16}=\frac{1\pm\sqrt{5}}{4}$

The answer is: $\frac{1\pm\sqrt{5}}{4}$

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Example Question #42 : Quadratic Formula

Solve for the roots, if any: y= -8x^2-x-1

Possible Answers:

$-\frac{1}{16}\pm\frac{i\sqrt{33}}{16}$

$-\frac{1}{8}\pm\frac{i\sqrt{33}}{8}$

$\textup{No solution.}$

$-\frac{1}{16}\pm\frac{i\sqrt{31}}{16}$

$-\frac{1}{4}\pm\frac{i\sqrt{31}}{4}$

Correct answer:

$-\frac{1}{16}\pm\frac{i\sqrt{31}}{16}$

Explanation:

This polynomial is in the standard form of a parabola, y=ax^2+bx+c .

Write the quadratic formula.

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Identify the variables.

a=-8, b=-1, c=-1

$x= \frac{-(-1)\pm\sqrt{(-1)^2-4(-8)(-1)}}{2(-8)}$

Simplify the quadratic.

$x= \frac{1\pm\sqrt{1-32}}{-16}=\frac{1\pm\sqrt{-31}}{-16}= \frac{1\pm i\sqrt{31}}{-16}$

Split the fraction.

The answer is: $-\frac{1}{16}\pm\frac{i\sqrt{31}}{16}$

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Example Question #501 : Intermediate Single Variable Algebra

Determine the roots, if possible: y=15x^2-x-1

Possible Answers:

$\frac{1\pm i\sqrt{31}}{15}$

$\textup{There are no possible roots.}$

$\frac{1\pm\sqrt{61}}{30}$

$\frac{1\pm i\sqrt{61}}{30}$

$\pm\frac{2 i\sqrt{15}}{15}$

Correct answer:

$\frac{1\pm\sqrt{61}}{30}$

Explanation:

Write the quadratic equation.

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

This equation will apply for polynomials in y=ax^2+bx+c format.

$y=15x^2-x-1 \rightarrow a=15, b=-1, c=-1$

Substitute the correct values into the quadratic formula.

$x= \frac{-(-1)\pm\sqrt{(-1)^2-4(15)(-1)}}{2(15)}$

Simplify this equation.

$x= \frac{1\pm\sqrt{1+60}}{30} = \frac{1\pm\sqrt{61}}{30}$

The answer is: $\frac{1\pm\sqrt{61}}{30}$

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Example Question #48 : Quadratic Formula

A ball is launched straight up from the ground at time t=0 , and the height of the ball from the ground at time is described by the function h(t)=-3x^2 +62x . Find the time at which the ball returns to the ground.

Possible Answers:

t=16

$t=\frac{67}{3}$

$t=\frac{62}{3}$

$t=\frac{60}{3}$

t=15

Correct answer:

$t=\frac{62}{3}$

Explanation:

In order to solve this problem, we think about how to translate what is asked of us in the problem into a mathematical notion. The problem asks us to find when the ball returns to the ground. We know that the ball is launched from the ground at time t=0 , and if we plug t=0 into our original function, we will find that at time t=0 , h(0)=(0)^2-8(0)=0 . Therefore, our "ground" is located at h(t)=0 , which is also the horizontal axis. This means that if we are looking for when our ball returns to the ground, mathematically we are looking for values of which make h(t)=0 . In other words, we are looking for the zeros (or horizontal intercepts) of our function.

Typically, the easiest way to find zeros is to factor, but in this situation factoring won't get us very far. So, we have to turn the quadratic formula. Remember for a function f(x) in the standard quadratic form, f(x)=ax^2+bx+c , you can find the zeros by plugging them into the quadratic formula, which is...

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

The first step in using the quadratic formula is to identify the proper , , and . In our case, a=-3 and b=62 , while c=0 . So, to solve for our zeros, we simply need to plug these numbers into the quadratic formula and solve for x.

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t=\frac{-(62)\pm \sqrt{(-62)^2-4(-3)(0)}}{2(-3)}$

$t=\frac{-62\pm \sqrt{3844}}{-6}$

$t=\frac{-62\pm 62}{-6}$

$t=\frac{0}{-6},\frac{-124}{-6}$

$t=0,\frac{62}{3}$

Since we know that t=0 is when our ball was launched, we know that $t=\frac{62}{3}$ must be when our ball returns to the ground.

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Example Question #502 : Intermediate Single Variable Algebra

Find the zeros of the function f(x) where

f(x)=6x^2+11x-35

Possible Answers:

$x=-\frac{7}{2},\frac{5}{3}$

$x=\frac{7}{3},\frac{5}{4}$

$x=\frac{7}{2},\frac{5}{3}$

$x=-\frac{7}{3},-\frac{5}{4}$

$x=\frac{1}{3},-\frac{1}{2}$

Correct answer:

$x=-\frac{7}{2},\frac{5}{3}$

Explanation:

Typically, the easiest way to find the zeros of a function is to factor. However, with this quadratic, which may not be particularly easy to factor by hand, quadratic formula may be easier. Recall that for a polynomial in the standard quadratic form, f(x)=ax^2+bx+c , the quadratic formula will give you all real zeros of the polynomial. First, we need to identify what , , and are in our original function. In this case, a=6 , b=11 , and c=-35 . Next, we need to set up the quadratic formula, plug in our values, and solve.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-11\pm \sqrt{(11)^2-4(6)(-35)}}{2(6)}$

$x=\frac{-11\pm \sqrt{121+840}}{12}$

$x=\frac{-11\pm \sqrt{961}}{12}$

$x=\frac{-11\pm 31}{12}$

$x=-\frac{42}{12},\frac{20}{12}$

$x=-\frac{7}{2},\frac{5}{3}$

Now we have our zeros!

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Example Question #43 : Quadratic Formula

Determine a root: y=-3x^2-x-1

Possible Answers:

$-\frac{1}{6}+\frac{i\sqrt{11}}{6}$

$-\frac{1}{3}+\frac{i\sqrt{11}}{3}$

$-\frac{1}{6}+\frac{2i\sqrt{6}}{3}$

$-\frac{1}{6}+\frac{i\sqrt{22}}{6}$

$-\frac{1}{2}+\frac{i\sqrt{11}}{2}$

Correct answer:

$-\frac{1}{6}+\frac{i\sqrt{11}}{6}$

Explanation:

Identify the coefficients given the equation in standard form:

y=ax^2+bx+c

a=-3, b=-1,c=-1

Write the quadratic formula.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the coefficients.

$x=\frac{-(-1)\pm \sqrt{(-1)^2-4(-3)(-1)}}{2(-3)}$

$x=\frac{1\pm \sqrt{1-12}}{-6} = \frac{1\pm \sqrt{-11}}{-6} = \frac{1\pm i\sqrt{11}}{-6}$

These two roots are complex and either is a valid answer.

The answer is: $-\frac{1}{6}+\frac{i\sqrt{11}}{6}$

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Example Question #191 : Solving Quadratic Equations

Determine a possible root for: y=x^2+10x-1

Possible Answers:

$5+3\sqrt{2}$

$\sqrt{13}-5$

$\sqrt{26}-5$

$\textup{None of the answers are possible roots.}$

$5-2\sqrt6$

Correct answer:

$\sqrt{26}-5$

Explanation:

Write the quadratic formula.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Identify and substitute the terms.

a=1,b=10, c=-1

$x=\frac{-10\pm \sqrt{10^2-4(1)(-1)}}{2(1)}=\frac{-10\pm \sqrt{100+4}}{2}$

$x=\frac{-10\pm \sqrt{104}}{2}$

Factor the radical using factors of perfect squares.

$\sqrt{104} = \sqrt4 \cdot \sqrt{26} = 2\sqrt{26}$

$x=\frac{-10\pm2 \sqrt{26}}{2} = -5\pm \sqrt{26}$

These two answers are possible roots.

The answer is: $\sqrt{26}-5$

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #331 : Quadratic Equations And Inequalities

Example Question #501 : Intermediate Single Variable Algebra

Example Question #44 : Quadratic Formula

Example Question #41 : Quadratic Formula

Example Question #42 : Quadratic Formula

Example Question #501 : Intermediate Single Variable Algebra

Example Question #48 : Quadratic Formula

Example Question #502 : Intermediate Single Variable Algebra

Example Question #43 : Quadratic Formula

Example Question #191 : Solving Quadratic Equations

All Algebra II Resources

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