12 squared = x(x+6)
144 = 2x +6x
144 = 8x
144/8 = 8x/8
18 = x
What is the answer to 12 squared = x(x+6) ?
12 answers
oh....it's false
12 squared=x(x+6)
144=x²+6x
x²+6x+9=153
(x+3)²=153
answers: x+3=3*sqrt(17)
or x+3=-3*sqrt(17)
12 squared=x(x+6)
144=x²+6x
x²+6x+9=153
(x+3)²=153
answers: x+3=3*sqrt(17)
or x+3=-3*sqrt(17)
OOPS, x(x+6) = x^2 + 6x
so x^2 + 6x - 144 = 0
x = (-3 ±√153)
so x^2 + 6x - 144 = 0
x = (-3 ±√153)
what do the question marks signify?
Oops! is right. I goofed. Please disregard my answer.
Hi:
Given 12^2 = x(x+6), we have 144 = x^2 + 6x. Now, setting one side equal to zero gives x^2 + 6x - 144 = 0. The quadratic is not factorable (do you know why?) so we use the quadratic formula to get,
x = 0.5[-6 +/- sqrt(36 - 4(-144))]
= 0.5[-6 +/- sqrt(612)]
= 0.5[-6 +/- 6sqrt(17)]
= -3 +/- 3sqrt(17)
= -3(1 +/- SQRT(17))
So your two solutions are
x = -3(1 + sqrt(17)) and
x = -3(1 - sqrt(17)).
Regards,
Rich B.
Given 12^2 = x(x+6), we have 144 = x^2 + 6x. Now, setting one side equal to zero gives x^2 + 6x - 144 = 0. The quadratic is not factorable (do you know why?) so we use the quadratic formula to get,
x = 0.5[-6 +/- sqrt(36 - 4(-144))]
= 0.5[-6 +/- sqrt(612)]
= 0.5[-6 +/- 6sqrt(17)]
= -3 +/- 3sqrt(17)
= -3(1 +/- SQRT(17))
So your two solutions are
x = -3(1 + sqrt(17)) and
x = -3(1 - sqrt(17)).
Regards,
Rich B.
you're right but saying x²+6x-144=0 is not factorable, it is a mistake
x²+6x-144=x²+6x+9-153=(x+3)²-153=(x+3+sqrt(153))*(x+3-sqrt(153))
so u can factorize it
x²+6x-144=x²+6x+9-153=(x+3)²-153=(x+3+sqrt(153))*(x+3-sqrt(153))
so u can factorize it
It is generally agreed in both the US and Canada that "factoring" is restricted to the set of rational numbers.
ok, thanks.
in france 'factoring'is restricted to 3 'identities' as a²-b²=(a-b)(a+b)
a²+2ab+b²=(a+b)² and a²-2ab+b²=(a-b)²...so diferents countries=diferents matematics lol
in france 'factoring'is restricted to 3 'identities' as a²-b²=(a-b)(a+b)
a²+2ab+b²=(a+b)² and a²-2ab+b²=(a-b)²...so diferents countries=diferents matematics lol
That makes very little sense.
You mean 10x^2 + 13x - 3 = (2x+3)(5x-1)
would not be acceptable in France??
I doubt that very much.
You mean 10x^2 + 13x - 3 = (2x+3)(5x-1)
would not be acceptable in France??
I doubt that very much.
sorry, but u say that 'factoring' is restricted to the set of rational numbers..and about the cumplex numbers???
mk-tintin:
Yes, of course. However, as another member writes, I assume one understands my meaning specifically as irreducible over the rationals, Q.
Rich B.
Yes, of course. However, as another member writes, I assume one understands my meaning specifically as irreducible over the rationals, Q.
Rich B.