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Precalculus : Solve a Quadratic Equation

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Example Questions

Example Question #3 : Solving Quadratic Equations

Solve the following quadratic equation:

3x^2-9x-84=0

Possible Answers:

x=-4,x=7

$x=\frac{3}{2},x=4$

x=1,x=3

x=-6,x=8

x=-2,x=5

Correct answer:

x=-4,x=7

Explanation:

When solving a quadratic equation, the first thing to look for is whether or not it can be factored, as this is most often the easiest and fastest method if the quadratic can in fact be factored. We can see that each of the terms in the given equation have a common factor of 3, so it will be easier to factor the quadratic if we first factor out the 3:

3x^2-9x-84=0

3(x^2-3x-28)=0

Now we're left with a polynomial where we need to find two numbers whose product is -28 and whose sum is -3. Thinking about the factors of 28, we can see that 4 and 7 will yield -3 if 7 is negative and 4 is positive, so we now have our factorization:

3(x-7)(x+4)=0

$x-7=0\rightarrow x=7$

$x+4=0\rightarrow x=-4$

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Example Question #1 : Solve A Quadratic Equation

Find the root(s) of the following quadratic equation?

y=x^2+9x+20

Possible Answers:

x=3

x={2,10}

x={4,5}

x={-5,-4}

x={-10,-2}

Correct answer:

x={-5,-4}

Explanation:

To find the roots of an equation in the form y=ax^2+bx+c , you use the quadratic formula

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

In our case, we have a=1, b=9, c=20 .

This gives us $\frac{(-9\pm\sqrt{9^2-4(1)(20)})}{2}$ which simplifies to

$\frac{(-9\pm\sqrt{81-80})}{2}=\frac{(-9\pm1)}{2}={-5,-4}$

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Example Question #5 : Solving Quadratic Equations

Given the function y=x^2-20x+51 , find a possible root for this quadratic.

Possible Answers:

Correct answer:

Explanation:

Factorize y=x^2-20x+51 and set this equation equal to zero.

y=x^2-20x+51=(x-17)(x-3)

(x-17)(x-3)=0

x=17,3

The answer is one of the possible choices.

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Example Question #11 : Solving Quadratic Equations

Solve the quadratic equation for .

(x-4)^2-16=0

Possible Answers:

None of the other answers.

Correct answer:

None of the other answers.

Explanation:

There are two solutions; x=0, x=8 .

We proceed as follows.

(x-4)^2-16=0

Add to both sides.

(x-4)^2=16

Take the square root of both sides, remember to introduce plus/minus on the right side since you are introducing a square root into your work.

$x-4 = \pm4$

Add to both sides.

x = 0, 8

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Example Question #12 : Solving Quadratic Equations

Solve the quadratic equation for .

4x^2 +5x -6 = 0

Use the quadratic formula.

Possible Answers:

$-\frac{5}{8} \pm \frac{1}{8}\sqrt{71}i$

4,0

$-2, \frac{3}{4}$

None of the other answers.

Correct answer:

$-2, \frac{3}{4}$

Explanation:

For any quadtratic equation of the form ax^2+bx+c=0 , the quadratic formula is

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

Plugging in our given values a = 4, b=5, c=-6 we have:

$x = \frac{-5 \pm \sqrt{25+96}}{8}$

$x = -\frac{5}{8} \pm \frac{\sqrt{121}}{8}$

$x= -\frac{5}{8} \pm \frac{11}{8}$

$x = -2, \frac{3}{4}$

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Example Question #6 : Solve A Quadratic Equation

Find the roots of the equation.

$2x^{2}-x-6=0$

Possible Answers:

$x=\pm3$

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

$x=3,\: \:x=-\frac{5}{3}$

$x=7,\: \:x=3$

$x=2,\: \:x=-7$

Correct answer:

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

Explanation:

Use either the quadratic formula or factoring to solve the quadratic equation.

Using factoring, we want to find which factors of six when multiplied with the factors of two and then added together result in negative one.

$2x^{2}-x-6=0$

(2x+3)(x-2)=0

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

Using the quadratic formula,

let

a=2, b=-1, c=-6

$x=\frac{1\pm\sqrt{1-4(2)(-6)}}{2(2)}$

$x=\frac{1\pm \sqrt{1+48}}{4}$

$\\x=\frac{1\pm \sqrt{49}}{4}=\frac{1\pm7}{4}\\ \\x=2, -\frac{3}{2}$

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Example Question #7 : Solve A Quadratic Equation

Solve f(x)=3x^2-4x+1 .

Possible Answers:

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

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

$x=0, \frac{1}{3}$

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

Correct answer:

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

Explanation:

To solve this equation, use trial and error to factor it. Since the leading coefficient is , there is only one way to get $(3\cdot1)$ , so that is helpful reminder. Once it's properly factored, you get: (3x-1)(x-1) . Then, set both of those expressions equal to to get your roots: $x=1, \frac{1}{3}$ .

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Precalculus : Solve a Quadratic Equation

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Solving Quadratic Equations

Example Question #1 : Solve A Quadratic Equation

Example Question #5 : Solving Quadratic Equations

Example Question #11 : Solving Quadratic Equations

Example Question #12 : Solving Quadratic Equations

Example Question #6 : Solve A Quadratic Equation

Example Question #7 : Solve A Quadratic Equation

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