Question

16x^2 + 24x + 9 = 0.. step by step?

Answer 1

i really need help. can you help me?

Answer 2

Factors of 9 are 9 and 1 or 3 and 3. (9 and 1 isn't going to work is it? So let's try 3 and 3.

So we will have (px + 3)(qx +3) =0

So multiply that out and you get

qpx^2 + 3(q+p)x + 9 =0

So qp =16 and 3(q+p)=24

Fortunately this points us at fours, does it not? But you can doi the substitution if you want.

And so,

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

So x=-3/4

Answer 3

Where did you get the q and the p ?

Answer 4

They are just used to represent the leading numbers that we didn't know yet. They could have been different but are the same because it is a squared function.
(4x+3)^2 = 0

Answer 5

I'm still confused, is there anyway you can explain it more?

Answer 6

OK - but I am eating my supper now and will do this for you in half-an-hour. ;-) Bear with me.

Answer 7

OK, back again...

The answer is of the form:
(p.x+d)(q.x+e) = 16x^2 + 24x +9

If we multiply that out we get:
p.q.x^2 + (q.d + p.e)x +e.d = 16x^2 + 24x +9

So p.q =16 and e.d = 9 and (q.d +p.e) = 24

As I said at the start, two numbers multiplied together making must be 9 and 1 or 3 and 3. Two numbers multiplied together making 16 must 1 and 16, or 2 and 8, or 4 and 4.

I chose p and e to be 3 and 3 because 9 and 1 didn't look very promising.

So plugging 3 and 3 into (q.d +p.e) = 24 we get
(3.q + 3.p) = 3(q+p)=24
And therefore, q + p = 8

Our choices for q and p are 1 and 16, or 2 and 8, or 4 and 4, and the only pair of those that add up to eight are 4 and 4.

So p = q = 4, and d = e = 3

Our original equation is
(p.x+d)(q.x+e) = 16x^2 + 24x +9

Just plug in the numbers...

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

If that is true one or other of the brackets must be equalt to zero too (in this case both of them are)

So 4x+3 =0, 4x=-3, and therefore x=-3/4

OK? :-)

Answer 8

THANK YOU (: