Help Find Solution Sets for Quadratic Equations

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Solve for x.

$x^{2}+9x+18=0$

Possible Answers:

x = 6, 3

x = –9, –2

x = –2

Cannot be factored by grouping

x = –6, –3

Correct answer:

x = –6, –3

Explanation:

$x^{2}+9x+18=0$

1) This is a relatively standard quadratic equation. List and add factors of 18.

1 + 18 = 19

2 + 9 = 11

3 + 6 = 9

$x^{2}+3x+6x+18=0$

2) Pull out common factors of each pair, "x" from the first and "6" from the second.

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

3) Factor again, pulling out "(x+3)" from both terms.

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

4) Set each term equal to zero and solve.

x + 3 = 0, x = –3

x + 6 = 0, x = –6

Solve for x.

$4x^{2}+8x+4=0$

Possible Answers:

x = 1

x = 4

x = –4

x = 2, 4

x = –1

Explanation:

$4x^{2}+8x+4=0$

1) After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

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

Since zero divided by four is still zero, only the left side of the equation changes.

2) Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

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

x(x+1)+1(x+1)=0

(The "1" was pulled out only to make the next factoring step clear.)

(x+1)(x+1)=0

x + 1 = 0, x = –1

Perfect Square:

$\sqrt{x^{2}}=\left | x\right |=x$

$\sqrt{1}=1$

$(x+1)^{2}=0$

x = –1

Solve for x.

$-15x+4x^{2}-4=0$

Possible Answers:

Cannot be factored by grouping

x = –1/4

x = 4, –1/4

x = –1, 1

x = –4, 4

Correct answer:

x = 4, –1/4

Explanation:

$-15x+4x^{2}-4=0$

1) Quadratics tend to be easier to follow when stated in order of descending degree. In other words, we need to rearrange the euqation.

$4x^{2}-15x-4=0$

2) No other simplification is possible, as there are no common factors between 15 and 4. Multiply the first coefficient by the final term and list off factors.

4 * –4 = –16

Factors of –16 include:

–1 + 16 = 15

1 + –16 = –15

3) Split up the middle term so that factoring by grouping is possible.

$4x^{2}+x-16x-4=0$

4) Factor by pulling the greatest common factor out of each pair of terms, "x" from the first and "-4" from the second.

x(4x+1)-4(4x+1)=0

5) Factor out the "4x+1" from both terms.

(x-4)(4x+1)=0

6) Set both parts equal to zero and solve.

x – 4 = 0, x = 4

4x + 1 = 0, x = –1/4

Solve for x.

7x^2-4x+13=0

Correct answer:

$x=\frac{2}{7}\pm \frac{i\sqrt{87}}{7}$

Explanation:

There are two ways to do this. One way involves using the quadratic formula. The quadratic formula is written below.

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

By looking at 7x^2-4x+13=0 , a = 7, b = –4, and c = 13. Plug these values into the quadratic equation to find x.

$x=\frac{-(-4)\pm \sqrt{(-4)^2-4(7)(13)}}{2(7)}$

$x=\frac{4\pm \sqrt{16-364}}{14}$

$x=\frac{4\pm \sqrt{-348}}{14}$

$x=\frac{4\pm \sqrt{(-1)(4)(87)}}{14}$

Note that $i=\sqrt{-1}$ .

$x=\frac{4\pm 2i\sqrt{(87)}}{14}$

Factor out the two, then cancel out that two and separate terms.

$x=\frac{2(2\pm i\sqrt{(87)})}{2(7)}$

$x=\frac{2}{7} \pm\frac{i\sqrt{87}}{7}$ This is our answer by the first merthod.

The other method to solve involves completing the square.

7x^2-4x+13=0

Subtract 13 to both sides.

7x^2-4x=-13

Divide 7 to both sides.

$x^2-\frac{4}{7}x=-\frac{13}{7}$

Take the –4/7 from the x-term, cut it in half to get –2/7. Square that –2/7 to get 4/49. Finally, add 4/49 to both sides

$x^2-\frac{4}{7}x+\frac{4}{49}=-\frac{13}{7}+\frac{4}{49}$

Factor the left hand side and simplify the right hand side.

$\left ( x-\frac{2}{7}\right )\left ( x-\frac{2}{7}\right )=-\frac{91}{49}+\frac{4}{49}$

$\left ( x-\frac{2}{7}\right )^2=-\frac{87}{49}$

Square root and add 2/7 to both sides.

$x=\frac{2}{7} \pm \sqrt{\frac{-87}{49}}$

Don't forget to write it in terms of 'i'.

$x=\frac{2}{7}\pm \frac{i\sqrt{87}}{7}$

Note that we should find the same answer by either method.

Billy is several years older than Johnny. Billy is one less than twice as old as Johnny, and their ages multiplied together make ninety-one. When will Billy be 1.5 times Johnny's age?

Possible Answers:

When Johnny is 7 and Billy is 13

When Johnny is 12 and Billy is 18

When Johnny is 2 and Billy is 3

When Johnny is 14 and Billy is 21

When Johnny is 4 and Billy is 6

Correct answer:

When Johnny is 12 and Billy is 18

Explanation:

1) Before we can figure out when Billy will be 1.5 times Johnny's age, we have to figure out their current ages. So let's define our variables in terms of the first part of the question.

B = Billy's age and J = Johnny's age

It's easier to solve if we put one variable in terms of the other. If Billy were just twice as old as Johnny, we could write his age as B = 2J.

But Billy isone less than twice as old as Johnny, so B = 2J – 1

2) We know that the two boys' ages multiply together to make ninety-one.

B * J = J(2J – 1) = 91

3) Now we have our factored quadratic. We just need to multiply it out and set everything equal to zero to begin.

J(2J-1)=91

$2J^{2}-J-91=0$

4) Now we need to factor back out. We start by multiplying the first coefficient by the final term and listing off the factors.

2 * –91 = –182

1 + –182 = –181

2 + –91 = –89

7 + –26 = –19

13 + –14 = –1

5) Split up the middle term in so that factoring by grouping is possible.

$2J^{2}-14J+13J-91=0$

6) Factor by grouping, pulling out "2J" from the first set of terms and "13" from the second.

2J(J-7)+13(J-7)=0

7) Factor out the "(J-7)" from both terms.

(2J+13)(J-7)=0

8) Set both parentheses equal to zero and solve.

2J + 13 = 0, J = –13/2

J – 7 = 0, J = 7

Clearly, only of the two solutions works, since Johnny's age has to be positive. Johnny is 7, therefore Billy is 2(7) – 1=13. But we're not done yet!

9) We need to figure out at what point Billy will 1.5 times Johnny's age. Guess and check would be a fairly efficient way to do this problem, but setting up an equation would be even faster. First, though, we need to figure out what our variable is. We know Billy's and Johnny's current ages; we just need to figure out their future ages. One variable is always better than two, so instead of using two different variables to represent their respective future ages, we'll use one variable to represent the number of years we have to add to each of their current ages in order to make Billy 1.5 times older than Johnny. Let's call that variable "x."

1.5(J + x) = B + x

We know the values of J and B, so we can go ahead and fill those in.

1.5(7 + x) = 13 + x

10) Then we solve for x algebraically, with inverse order of operations.

10.5 + 1.5x = 13 + x

0.5x = 2.5

x = 5

J = 7 + 5 = 12

B = 13 + 5 = 18

Find all of the solutions to the following quadratic equation:

$2x^{2} - 7x + 4 = 0$

Correct answer:

$x = \frac{7 \pm \sqrt{17}}{4}$

Explanation:

This requires the use of the quadratic formula. Recall that:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ for $ax^{2} + bx + c = 0$ .

For this problem, $a=2,b=-7,\ and\ c=4$ .

So,

$x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4(2)(4})}{2(2)}$ .

$x = \frac{7 \pm \sqrt{49-32}}{4}$ .

Therefore, the two solutions are:

$x = \frac{7 \pm \sqrt{17}}{4}$

Solve for.

(x-3)(x+3) =5x+5

Correct answer:

$-2 \;or\; 7$

Explanation:

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

(x-3)(x+3) =5x+5

First: x*x=x^2

Inside: -3*x=-3x

Outside: x*3=3x

Last: -3*3=-9

$x^{2}-3x+3x-9 =5x+5$

x^2-9=5x+5

$x^{2}-5x-14 =0$

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product; these numbers are -7,2 .

-7*2=-14 and -7+2=-5

(x-7)(x+2) =0

x-7 =0

x=7

x+2 =0

x=-2

The solution set is $\{-2, 7\}$ .

Solve for : 3x(x+4)=x+20

Correct answer:

$\left \{ -5, \frac{4}{3}\right \}$

Explanation:

Eliminate parentheses, then write this quadratic equation in standard form, with all nonzero terms on one side:

3x(x+4)=x+20

$3x \cdot x +3x \cdot 4=x+20$

$3x^{2} +12x=x+20$

$3x^{2} +12x-x-20 =0$

$3x^{2} +11x-20 =0$

Now factor the quadratic expression using the -method - split the middle term into two terms whose coefficients add up to 11 and have product $3 \cdot (-20) = -60$ . These numbers are 15,-4

$3x^{2} -4x+15x-20 =0$

x (3x -4) + 5(3x -4)=0

(x + 5)(3x -4) =0

Set each factor to 0:

x+5 = 0

x = -5

3x-4 = 0

3x=4

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

$Solve\;for\;x,\;(x-3)^2 =9$

Correct answer:

$\fn_cm 0,6$

$Solve \;for\;p,(p-5)(p+3)=0$

Correct answer:

5,-3

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Help Find Solution Sets for Quadratic Equations

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Help Find Solution Sets for Quadratic Equations

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