Question

how to do solve this problem by using the quadratic formula (x+18)^2= 120

Answer 1

you how to squaring ?

Answer 2

Expand (x+18)^2 and set the equation equal to zero then apply the quadratic formula. If you are having issues with the application, note that a is the coefficient of x^2, b is the coefficient of x and c is the remaining term (no x). Let us know what you have so far and may be having issues with.

Answer 3

It woudl be \[x ^{2}+36x+324=120\], correct?

Answer 4

Then subtract 120 to both sides and I would get \[x ^{2}+36x+204=0\]

Answer 5

That is what I got.

Answer 6

Then I would apply to the quadratic formula which would be \[x= -(36) \pm \sqrt{(36)^{2}-4(1)(204)}\] over 2(1)

Answer 7

Yes, that's correct.

Answer 8

I got x to equal \[-18\pm 2\sqrt{30}\]. Is this what you got?

Answer 9

If you didn't leave it in the exact form x = -7.05 and -28.95

Answer 10

No, can you explain it to me how you got that answer please. I'm stuck on this part

Answer 11

Sure.
First, I would do the 'stuff' inside the radical. We have 36^2 and (204*4). Since it is 36^2 - (204*4), we get 1296 - 816 = 480.
At this point we have: \[\frac{ -36\pm \sqrt{480} }{ 2 }\].

I simplified the radical by setting \[\sqrt{480} = \sqrt{4} * \sqrt{120} =2 *\sqrt{120} = 2\sqrt{4} * \sqrt{30} = 4\sqrt{30}\]
If you don't know how to do the above, you can simply break radicals into 'smaller' ones using their factors. To simplify the radical your goal should be to use factors that have square roots, for example 4 was used twice. Sqrt(30) cannot be further reduced because its factors (1,2,3,5,6,10,15,30) contain zero numbers which have an exact square root, like 4 does. So, we then divide boht sides by 2 and get: \[-18\pm 2\sqrt{30}\]

Answer 12

Oh ok, I get it. but you know where you got \[2\sqrt{4} * \sqrt{30}\] what happen to the 2 in front of the square root of 4?

Answer 13

Because \(\sqrt4=2\)