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Algebra II : Quadratic Formula

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Example Questions

Example Question #11 : Quadratic Formula

Solve for . Use the quadratic formula to find your solution. Use a calculator to estimate the value to the closest hundredth.

5x^2 - 17x + 12 = 0

Possible Answers:

17.71 and -6.17

No solution

8.13 and 4.21

and 2.4

1.52 and 3.52

Correct answer:

and 2.4

Explanation:

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

ax^2+bx+c=0

5x^2 - 17x + 12 = 0

a = 5

b = -17

c = 12

Substituting these values into the equation, you get:

$\frac{-(-17) \pm \sqrt{(-17)^2 - 4(5)(12)}}{2(5)}$

$\frac{17 \pm \sqrt{289 - 240}}{10}$

$\frac{17 \pm \sqrt{49}}{10}$

$\frac{17 \pm 7}{10}$

Separate this expression into two fractions and simplify to determine the final values.

$\frac{17 + 7}{10} = \frac{24}{10} = 2.4$

$\frac{17 - 7}{10} = \frac{10}{10} = 1$

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Example Question #471 : Intermediate Single Variable Algebra

Solve for . Use the quadratic formula to find your solution. Use a calculator to estimate the value to the closest hundredth.

3x^2 +8x - 11 = 0

Possible Answers:

and -3.67

5.28 and -2.34

9.87 and -6.87

No solution

-3.28 and 2.83

Correct answer:

and -3.67

Explanation:

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

ax^2+bx+c=0

3x^2 +8x - 11 = 0

a = 3

b = 8

c = -11

Substituting these values into the equation, you get:

$\frac{-(8) \pm \sqrt{(8)^2 - 4(3)(-11)}}{2(3)}$

$\frac{-8 \pm \sqrt{64 - (-132)}}{6}$

$\frac{-8 \pm \sqrt{64 +132}}{6}$

$\frac{-8 \pm \sqrt{196}}{6}$

$\frac{-8 \pm 14}{6}$

Separate this expression into two fractions and simplify to determine the final values.

$\frac{-8 + 14}{6} = \frac{6}{6} = 1$

$\frac{-8 - 14}{6} = \frac{-22}{6} = -3.67$

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Example Question #471 : Intermediate Single Variable Algebra

Solve the quadratic equation with the quadratic formula.

$2x^{2}+3x-1=0$

Possible Answers:

$x=\frac{-3\pm\sqrt{1}}{4}$

$x=\frac{-3\pm\sqrt{17}}{4}$

$x=\frac{-3\pm\sqrt{17}}{2}$

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

$x=\frac{3\pm\sqrt{17}}{4}$

Correct answer:

$x=\frac{-3\pm\sqrt{17}}{4}$

Explanation:

Based on the quadratic equation:

$ax^{2}+bx+c=0$ ,

a=2 , b=3 , and c=-1

Given the quadratic formula:

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

We have:

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

Simplfying,

$x=\frac{-3\pm\sqrt{17}}{4}$

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Example Question #473 : Intermediate Single Variable Algebra

Use the quadratic equation to solve x^2-2x+1

Possible Answers:

x=1

x=2

x=-2

x=-1

Correct answer:

x=1

Explanation:

We use the quadratic equation to solve for x. The quadratic equation is:

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

In our case a=1, b=-2, c=1

Substituting these values into the quadratic equation we get:

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

$x=\frac{2+\sqrt{(4-4}}{2}, x=\frac{2-\sqrt{4-4}}{2}$

$x=\frac{2+0}{2}, x=\frac{2-0}{2}$

x=1, x=1

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Example Question #15 : Quadratic Formula

The height of a kicked soccer ball can be modeled with the equation

H(x)=-4.9x^2 +15x +0.5 ,

where the height is given in meters and is the time in seconds. At what time(s) will the ball be 2 meters off the ground?

Possible Answers:

x=0.412 seconds

x=1.244 seconds

x=0.104 seconds

x=2.958 seconds

x=1.244 seconds

x=0.104 seconds

Correct answer:

x=0.104 seconds

x=2.958 seconds

Explanation:

Set up the equation to solve for the time when the height is at 2 meters:

2=-4.9x^2+15x-0.5

Now put the equation into quadratic form ax^2+bx+c=0 so that we can solve it using the quadratic formula

-4.9x^2+15x-1.5=0 .

The quadratic equation is

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

where a=-4.9 , b=15 , and c=-1.5 .

Solving for gives us two possible values,

x=0.104 seconds

x=2.958 seconds.

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Example Question #11 : Quadratic Formula

Solve for .

x^2+4x+5=0

Possible Answers:

$\textup{Answer is not present}$

x=0, -4

x=2, -2

x=-2

Correct answer:

$\textup{Answer is not present}$

Explanation:

When applying the quadratic formula, the discriminant (portion under the square root) is negative and so there are no real roots of the equation shown.

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Example Question #475 : Intermediate Single Variable Algebra

Solve for x

(x+2)^2+2x+2=0

Possible Answers:

x=-9, -3

$x=3\pm\sqrt{3}$

x=-1, -5

$x=-3\pm{\sqrt{3}}$

Correct answer:

$x=-3\pm{\sqrt{3}}$

Explanation:

Once the square is multiplied out and the equation simplified, it yields x^2+6x+6=0 , a good time for the quadratic formula, $x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$ where a, b, c are the coefficients of the polynominal in descending order. Plug in a=1, b=6, c=6, and it yields $-3\pm{\frac{\sqrt{12}}{2}}$ , multiply out the square root and it yields $-3\pm{\sqrt{3}}$ .

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Example Question #476 : Intermediate Single Variable Algebra

Using the quadratic equation, find the roots of the following expression.

2x^2+5x - 12

Possible Answers:

x=-4

x=-1.5

x = -4

$x = \frac{2}{3}$

x =-4

x=1.5

x=4

x = 1.5

No real solutions

Correct answer:

x =-4

x=1.5

Explanation:

To find the roots of the quadratic expression, we must use the quadratic equation

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

Plugging in our values for , , and (, , and , respectively) we get the equation:

$x = \frac{-5\pm\sqrt{25-4(2)(-12)}}{2\cdot2}$

First, let's simplify the radical:

$x=\frac{-5\pm\sqrt{25+96}}{4}$

which becomes

$x=\frac{-5\pm\sqrt{121}}{4}$

$x=\frac{-5\pm11}{4}$

Now that we've simplified the radical, we need to solve for both solutions:

$x_1=\frac{-5+11}{4}=\frac{6}{4}=\frac{3}{2}=1.5$

and

$x_2=\frac{-5-11}{4}=\frac{-16}{4}=-4$

Therefore, the roots of this quadratic expression are x=1.5 and x=-4 .

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Example Question #311 : Quadratic Equations And Inequalities

Find the roots of the following equation using the quadratic formula:

f(x) = 3x^2 + 8x - 3

Express in simplest form.

Possible Answers:

$x = \frac{-8\pm10}{6}$

$x = \frac{-4\pm5}{3}$

$x = \frac{-8\pm2\sqrt{7}}{6}$

$x = \frac{-4\pm\sqrt{7}}{3}$

$x = \frac{-8\pm\sqrt{28}}{6}$

Correct answer:

$x = \frac{-4\pm5}{3}$

Explanation:

Remember the quadratic equation. For any quadratic polynomial, ax^2 + bx + c , the roots of the function are given by:

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

In this situation, we have 3x^2 + 8x -3 , so a = 3, b = 8, c = -3 .

Substituting into the formula, we get the roots at:

$x = \frac{-8\pm\sqrt{8^2-4(3\cdot-3)}}{2\cdot3}$

Simplifying gives us:

$x = \frac{-8\pm\sqrt{64+36}}{6} = \frac{-8\pm\sqrt{100}}{6} = \frac{-8\pm10}{6} = \frac{-4\pm5}{3}$ .

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Example Question #3 : How To Use The Quadratic Function

Which of the following is the correct solution when x^2-10x-11=0 is solved using the quadratic equation?

Possible Answers:

$\frac{10\pm \sqrt{56}}{2}$

$\frac{10\pm 12}{2}$

$10\pm 12$

$\frac{-10\pm 12}{2}$

Correct answer:

$\frac{10\pm 12}{2}$

Explanation:

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-11)}}{2(1)}$

$=\frac{10\pm \sqrt{100+44}}{2}$

$=\frac{10\pm \sqrt{144}}{2}$

$=\frac{10\pm 12}{2}$

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Algebra II : Quadratic Formula

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Quadratic Formula

Example Question #471 : Intermediate Single Variable Algebra

Example Question #471 : Intermediate Single Variable Algebra

Example Question #473 : Intermediate Single Variable Algebra

Example Question #15 : Quadratic Formula

Example Question #11 : Quadratic Formula

Example Question #475 : Intermediate Single Variable Algebra

Example Question #476 : Intermediate Single Variable Algebra

Example Question #311 : Quadratic Equations And Inequalities

Example Question #3 : How To Use The Quadratic Function

All Algebra II Resources

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