I have some questions I would like a step-by-step answer of.?

Question

1) x^2 = 120 - 7x. Find the value of 'x'.

2) Find the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and15m respectively and the angle between the first two sides is a right angle.

Answer 1

1) x^2 = 120 - 7x
x^2 + 7x - 120 = 0 {add 7x, subtract 120, both sides}

You need to find two factors of 120, one of them being negative, that add to 7.

-8 and 15 seem to work

(x - 8)(x + 15) = 0 {factor the quadratic}
x = 8, x = -15 {set each binomial to zero and solve for x}

2) Find the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and15m respectively and the angle between the first two sides is a right angle.

Don't quite see how to do this one - is all the information here?

Answer 2

OK, #1 first:

Step one, get all terms on one side of the equation, and zero on the other side.

x^2 + 7x - 120 = 120 - 120 - 7x + 7x

or...

x^2 + 7x - 120 = 0

Now, the problem can be factored. There are two numbers whose sum is +7, but whose product is -120, and we must find them. Turns out that +15 and -8 fill the bill, so the factored version is:

( x + 15 ) ( x - 8 ) = 0

So, one or the other of these factors, ( x + 15 ) or ( x - 8 ) must be zero in order to maintain the equality. This can happen if x = 8, or if x = -15.

So, x can be +8 or -15. This gives us the following intercepts on the y axis:

(8,0) and (-15,0)


As far as your second question goes... it's too late at night here to answer it.... my eyes are starting to cross!

Good luck!

Answer 3

(1)
x^2 = 120 - 7x
x^2 + 7x = 120
x^2 + 7x - 120 = 0
(x + 15)(x - 8) = 0

x +15 = 0 [OR] x -8 = 0
x = -15 [OR] x = 8

(2) Can't help you there...yet.

Answer 4

I don't recall the "math" involved, but the answer is 8. It is DEFINITELY NOT 1800.

8^2 = 120 -7*8
64 = 120 - 56
64=64

Answer 5

x^2=120-7x
x^2+7x=120
x+7x=120^2
8x=14400
x=14400/8
x=1800