Question

how do i solve this im so confused @.@ Solve the following equation: x^2 - 14x + 49 = 0
{7}
{-7}
{-7, 7}
{ 7,-7

Answer 1

You have a number of options.

Because this is a multiple-guess question, you could simply try out the answers and see which one works. For example, (7)^2 - 14(7) + 49 = 49 + 49 - 98 = 0, so 7 is a solution. How about -7? (-7)^2 - 14(-7) + 49 = 49 + 98 + 49 = 196, so -7 is not a solution.

Now, that works, but only if you have a short menu of answers from which to choose. Better to know how to solve it!

\[x^2-14+49 = 0\]Let's try factoring. What are two numbers that when multiplied give us 49, and when added, give us -14?

Answer 2

ill be honest just use mathway.com it will help you out alot

Answer 3

@chancehenninger Not when she's taking a test, it won't.

Answer 4

-7?

Answer 5

@EmmaH are you there?

Answer 6

my teachers always said its not cheating its using your resources

Answer 7

Right! So we can write our equation as \[(x-7)(x-7) = 0\]because \[(x-7)(x-7) = x^2-7x-7x+49 = x^2-14x+49\]

Now can you find the values of \(x\) that make \[(x-7)(x-7) = 0\] true?

Answer 8

your just confusing her even more

Answer 9

@chancehenninger Who said anything about cheating. I'm making sure Emma can solve the problem herself, so that she is her own best resource.

Answer 10

If you have a product of two numbers = 0, one or both of those numbers must be 0. For \[(x-7)(x-7) = 0 \rightarrow (x-7) = 0\]\[x-7 = 0\]\[x=7\]
So both solutions to the quadratic equation are x = 7.

If you get an equation that you can't or don't want to factor, you can use the quadratic formula to find the solutions of \(ax^2+bx+c=0\)
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]so long as \(a\ne0\)

In this case, we have \(a=1,b=-14,c=49\):
\[x=\frac{-(-14)\pm\sqrt{(-14)^2-4(1)(49)}}{2(1)} = \frac{14\pm\sqrt{196-196}}{2} = 7\pm 0 = 7 \]

Answer 11

ok im sorry but (x-7) = 0\]\[x-7 = 0\]\[x=7\]
is just a bunch o =f symnbols to me it makes no sense

Answer 12

Well, if I tell you that I have two numbers, and I multiply them together, and the answer is 0, what can you tell me about those numbers?

Answer 13

all im seing is atuff like this = 0\]

Answer 14

Oh, it isn't displaying your page properly :-( Try refreshing the page. Here, I'll attach pictures of the posts, properly displayed:

Answer 15

All right, then factoring is probably the only approach for you now. Do you see how it produced the answer?

Answer 16

can you show me how you got it from the beginnign steps im understanding the last steps but not the furst couple

Answer 17

The factoring part?

I'll do another one. This one will have two different solutions instead of one.

x^2 + 3x + 2 = 0 (I won't do the formatting this time)

To factor that, we need to find 2 numbers that when we multiply them give us 2, but when we add them give us 3. Here's why:

(x+a)(x+b) = x^2 + ax + bx + ab = x^2 + (a+b)x + ab
Now if we look at our equation we are trying to factor, we can see that those two look identical so long as ab = 2 and (a+b) = 3, right?

Answer 18

I'll take that as a yes :-)

1*2 = 2
1+2 = 3

So we can write our factored equation like this:

x^2 + 3x + 2 = (x+1)(x+2) = 0

Now, for that to be true, there are two possible values of x:

(x+1) = 0
(x+2) = 0

and those give us x = -1, x = -2 as our solutions.

Let's check them:

(-1)^2 + 3(-1) + 2 = 1 -3 + 2 = 0
(-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0