Algebra - How to find the solution to a quadratic equation

Find all of the solutions to the following quadratic equation:

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

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

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

$x = \frac{-7 \pm \sqrt{33}}{4}$

$x = \frac{7 \pm \sqrt{17}}{2}$

None of the above

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:

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

Find the roots of the following equation.

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

$x=\frac{-6\pm \sqrt{30}}{3}$

$x=\frac{-2\pm \sqrt{11}}{4}$

$x=\frac{7\pm \sqrt{2}}{3}$

Explanation

Use the quadratic formula to solve the equation.

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

Plug in these values and solve.

$x=\frac{-12\pm \sqrt{12^{2}-4(3)(2)}}{2(3)}$

$x=\frac{-12\pm \sqrt{144-24}}{6}$

$x=\frac{-12\pm \sqrt{120}}{6}$

$x=\frac{-6\pm \sqrt{30}}{3}$

Solve for x.

$x^{2}-7x+10=0$

x = 5, 2

x = –4, –3

x = –5, –2

x = 4, 3

x = 5

Explanation

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

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

$x^{2}-2x-5x+10=0$

Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

Now pull out the common factor, the "(x-2)," from both terms.

Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

Solve for .

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

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

$x=2\ \textup{or}\ x=-2$

Explanation

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.

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

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

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

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

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

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

Solve for x.

$x^{2}-5x+10=-13x-6$

x = –4

x = –4, 4

x = 5, 2

x = 2

x = –5, –2

Explanation

$x^{2}-5x+9=-13x-7$

First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

$x^{2}+8x+16=0$

There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

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

Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

Pull out the "(x+4)" to wind up with:

Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., $px^{2}+qx+k=0$ ), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses.

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

$\sqrt{16}=4$

$(x+4)^{2}=0$

And x, once again, is equal to –4.

Solve for x.

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

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

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

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

No solution

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 , 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.

Subtract 13 to both sides.

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.

Solve for $\small x$ .

$\small x^2+6x+8=0$

$\small x=-2\ or\ x=-4$

$\small x=2\ or\ x=4$

$\small x=2\ or\ x=-4$

$\small x=-2\ or\ x=4$

Explanation

$\small x^2+6x+8=0$

Solve by factoring. We need to find two factors that multiply to eight and add to six.

$\small 2*4=8\ \text{and}\ 2+4=6$

$\small x^2+6x+8=(x+2)(x+4)=0$

One of these factors must equal zero in order for the equation to be true.

$\small x+2=0\ or\ x+4=0$

$\small x=-2\ or\ x=-4$

Solve for x.

$2x^{2}+15x+25=0$

x = –5/2, –5

x = –5, 5

x = –5

x = –2/5, –5

x = –2/3, –3

Explanation

$2x^{2}+15x+25=0$

The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

Split up the middle term to make factoring by grouping possible.

$2x^{2}+10x+5x+25=0$

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

Pull out the common factors from both groups, "2x" from the first and "5" from the second.

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

Set each parenthetical expression equal to zero and solve.

2x + 5 = 0, x = –5/2

x + 5 = 0, x = –5

Find the solutions to this quadratic equation:

$x=-6\hspace{1mm}and\hspace{1mm}x=2$

$x=-6\hspace{1mm}and\hspace{1mm}x=-2$

$x=6\hspace{1mm}and\hspace{1mm}x=2$

$x=6\hspace{1mm}and\hspace{1mm}x=-2$

None of the other answers.

Explanation

Put the quadratic in standard form:

Factor:

$Solutions=-6\hspace{1mm}and\hspace{1mm}2$

An easy way to factor (and do so with less trial and error) is to think of what two numbers could multiply to equal "c", but add to equal "b". These letters come from the designations in the standard form of a quadratic equation: . As you can see the product of -6 and 2 is -12 and they both add to 4.

Solve for x.

x = –8, –2

x = 6, 4

x = 4

Cannot be factored by grouping.

x = –6, –4

Explanation

Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.

$x^{2}+4x+6x+24=8$