Quadratic Formula - Algebra 2

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Question

Use the quadratic formula to solve for . Use a calculator to estimate the value to the closest hundredth.

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-14 \pm \sqrt{(14)^2 - 4(3)(9)}}{2(3)}$

$\frac{-14 \pm \sqrt{196 - 108}}{6}$

$\frac{-14 \pm \sqrt{88}}{6}$

Use a calculator to determine the final values.

$\frac{14+\sqrt{88}}{6}=3.90$

$\frac{14-\sqrt{88}}{6}=0.77$

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Question

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

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-5 \pm \sqrt{(5)^2 - 4(1)(-5)}}{2(1)}$

$\frac{-5 \pm \sqrt{25 +20)}}{2}$

$\frac{-5 \pm \sqrt{45}}{2}$

Use a calculator to determine the final values.

$\frac{-5+\sqrt{45}}{2}=0.85$

$\frac{-5-\sqrt{45}}{2}=-5.85$

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Question

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

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

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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Question

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

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

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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Question

Use the quadratic formula to solve for x:

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Answer

To solve this problem, you must first rewrite the equation into form (quadratic form).

After this you plug the numbers into the following quadratic equation:

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

Which upon doing you get:

$x=\frac{-4\pm \sqrt{16-8}}{2}$

This simplifies to:

$x=-2\pm\sqrt{2}$

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Question

Solve the equation using the quadratic formula:

$\small 0=2x^2+x-7$

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Answer

The quadratic formula:

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

$\small a=2; b=-1; c=7$

$\small x=\frac{-1\pm \sqrt{1^2-(4)(2)(-7)}}{(2)(2)}=\frac{-1\pm \sqrt{1+56}}{4}=\frac{-1\pm \sqrt{57}}{4}$

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Question

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

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}$

$\frac{4 \pm \sqrt{16 - 20}}{2}$

$\frac{4 \pm \sqrt{-4}}{2}$

Since you have to take the square root of a negative number, this means that there are no real solutions.

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Question

Solve for .

$x^{2}-4x-4=0$

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Answer

Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

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

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

$x=\frac{4\pm \sqrt{16+16}}{2}$

$x=\frac{4\pm \sqrt{32}}{2}$

$x=\frac{4\pm \sqrt{16\cdot 2}}{2}=\frac{4\pm \sqrt{16}\cdot \sqrt{2}}{2}=\frac{4\pm4\sqrt{2}}{2}$

$x=2\pm2\sqrt{2}$

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Question

A baseball that is thrown in the air follows a trajectory of , where is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

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Answer

To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then

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

Plugging in our values:

$t= \frac{-12\pm\sqrt{(12^2-4(-4)(6)}}{2(-4)}=\frac{-12\pm \sqrt{240}}{-8}= \frac{12\pm15.5}{8}$

Therefore or and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

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Question

Solve for :

$x^{2}+12x+35=0$

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Answer

To factor this equation, first find two numbers that multiply to 35 and sum to 12. These numbers are 5 and 7. Split up 12x using these two coefficients:

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Question

The product of two consective positive odd integers is 143. Find both integers.

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Answer

If is one odd number, then the next odd number is . If their product is 143, then the following equation is true.

Distribute into the parenthesis.

Subtract 143 from both sides.

$n^{2} + 2n - 143 = 0$

This can be solved by factoring, or by the quadratic equation. We will use the latter.

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

$n=\frac{-2 \pm \sqrt{4-(4(1)(-143)}}{2} = \frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}$

$n=\frac{22}{2}\ or\ n=\frac{-26}{2}$

$n=11\ or\ n=-13$

We are told that both integers are positive, so .

The other integer is .

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Question

Which of the following is the correct solution when is solved using the quadratic equation?

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Answer

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

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

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

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

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Question

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

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Answer

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-5)}}{2(1)}$

$\frac{7 \pm \sqrt{49 - (-20)}}{2}$

$\frac{7 \pm \sqrt{49 +20}}{2}$

$\frac{7 \pm \sqrt{69}}{2}$

Use a calculator to determine the final values.

$\frac{7+\sqrt{69}}{2}=7.65$

$\frac{7-\sqrt{69}}{2}=-0.65$

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Question

Use the quadratic formula to solve the following equation:

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Answer

Since the equation is already in standard form, we can compare it to the general version of standard form to find a, b, and c.

By pattern matching it is clear that a = 2, b = 4, and c = -3. Now we can substitute into the quadratic formula.

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

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

Recall that when subtracting a negative number \[as in - 4(2)(-3)\], the result is addition:

$x =\frac{-4 \pm\sqrt{16 + 24}}{4}$

Combine terms and simplify the radical:

$x =\frac{-4 \pm\sqrt{40}}{4}$

$x =\frac{-4 \pm\sqrt{4}*\sqrt{10}}{4}$

$x =\frac{-4 \pm\2\sqrt{10}}{4}$

Now eliminate common factors to reduce to lowest terms.

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

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

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Question

Solve for x

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Answer

Once the square is multiplied out and the equation simplified, it yields , 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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Question

Solve for .

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Answer

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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Question

Solve the quadratic equation with the quadratic formula.

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

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Answer

Based on the quadratic equation:

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

, , and

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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Question

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

,

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?

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Answer

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

Now put the equation into quadratic form so that we can solve it using the quadratic formula

.

The quadratic equation is

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

where , , and .

Solving for gives us two possible values,

seconds

or

seconds.

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Question

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

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Answer

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

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

or

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

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 and .

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Question

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

Express in simplest form.

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Answer

Remember the quadratic equation. For any quadratic polynomial, , the roots of the function are given by:

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

In this situation, we have , so .

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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