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ACT Math : Quadratic Equations

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Example Question #581 : Algebra

Solve for :

x^2 + 4x + 24 = 6 - 5x

Round to the nearest hundredth.

Possible Answers:

$4.44\textup{ and}-2.41$

$6\textup{ and }3$

$\textup{No solution}$

$-4.44\textup{ and }2.41$

$-6\textup{ and}-3$

Correct answer:

$-6\textup{ and}-3$

Explanation:

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify x^2 + 4x + 24 = 6 - 5x into:

x^2 + 9x + 18 = 0

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can. Factor the quadratic expression:

(x+3)(x+6)=0

Now, remember that you merely need to set each group equal to . This gives you the two values for :

x+3=0 ; therefore x=-3

Likewise, for the other group, x=-6

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Example Question #582 : Algebra

Solve for :

2x^2 + 19x + 10 = 4 - x^2

Possible Answers:

$-6\textup{ and }\frac{-1}{3}$

$\textup{No solution}$

$\frac{2}{5}\textup{ and }\frac{4}{5}$

$3\textup{ and }6$

$\frac{1 + \sqrt{6}}{3}\textup{ and }\frac{1 - \sqrt{6}}{3}$

Correct answer:

$-6\textup{ and }\frac{-1}{3}$

Explanation:

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify 2x^2 + 19x + 10 = 4 - x^2 into:

3x^2 + 19x + 6 = 0

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can, though we are sometimes a bit intimidated by x^2 terms that have a coefficient like this. Factor the quadratic expression:

(3x+1)(x+6)=0

If you FOIL this out, you will see your original equation.

Now, remember that you merely need to set each group equal to . This gives you the two values for :

3x+1=0

3x=-1

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

For the other group, you get x=-6 .

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Example Question #583 : Algebra

Solve for :

x^2 + 16x - 5 = 10 - 4x

Possible Answers:

$\frac{-20 \pm \sqrt{14}}{2}$

$\textup{No solution}$

$-10\pm\sqrt{115}$

$-30\textup{ and }30$

$\frac{-20 \pm \sqrt{30}}{2}$

Correct answer:

$-10\pm\sqrt{115}$

Explanation:

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify x^2 + 16x - 5 = 10 - 4x into:

x^2 + 20x - 15 = 0

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it cannot be easily factored. Therefore, you should use the quadratic formula. Recall that its general form is:

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

For our data, a=1 , b=20 , and c=-15 .

Thus, we have:

$\frac{-20\pm\sqrt{20^2-(4*1*-15)}}{2*1}$

Simplifying, this is:

$\frac{-20\pm\sqrt{400-(-60)}}{2}$

$\frac{-20\pm\sqrt{400+60}}{2}$

$\frac{-20\pm\sqrt{460}}{2}$

Now, simplify.

$\frac{-20\pm\sqrt{2*2*5*23}}{2}=\frac{-20\pm2\sqrt{115}}{2}=-10\pm\sqrt{115}$

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Example Question #584 : Algebra

Solve for :

7x^2 + 3x +25 = 4x - 10

Possible Answers:

$-12\textup{ and }4$

$-8.5\textup{ and}-3.5$

$\frac{-5 \pm \sqrt{31}}{3}$

$\frac{1 \pm \sqrt{491}}{-2}$

$\textup{No real solution}$

Correct answer:

$\textup{No real solution}$

Explanation:

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify 7x^2 + 3x +25 = 4x - 10 into:

7x^2 -x +35 = 0

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

For our data, a=7 , b=-1 , and c=35 .

Thus, we have:

$\frac{-(-1)\pm\sqrt{(-1)^2-(4*7*35)}}{2*-1}$

Simplifying, this is:

$\frac{1\pm\sqrt{1-(980)}}{-2}$

Since 1-980 is negative, you know that there is no real solution (given the problems arising from the negative square root)!

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Example Question #585 : Algebra

Solve: $x^{2}-5x+6$

Possible Answers:

and

None of the other answers

and

Correct answer:

and

Explanation:

To solve, we must set it equal to zero. The above expression is of the form $ax^{2}+bx+c$ , so we can use the quadratic formula:

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

to solve for the roots which are and

We can check by plugging the roots into the expression and making sure that it equals zero.

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ACT Math : Quadratic Equations

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #581 : Algebra

Example Question #582 : Algebra

Example Question #583 : Algebra

Example Question #584 : Algebra

Example Question #585 : Algebra

All ACT Math Resources

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