solve the equation and show work...the exponent threw me off
2007-09-18 17:52:38 · 11 answers · asked by arod_69 1 in Science & Mathematics ➔ Mathematics
An equation requires an = sign.
Assume equation is:-
(x - 6)² + x - 6 = 0
x² - 12x + 36 + x - 6 = 0
x² - 11x + 30 = 0
(x - 6)(x - 5) = 0
x = 6 , x = 5
2007-09-18 21:07:44 · answer #1 · answered by Como 7 · 2⤊ 1⤋
x-6[ 1 +(x-6)]=0
(x-6)(x-5)
x=6 and x=5
2007-09-18 18:00:16 · answer #2 · answered by Anonymous · 1⤊ 1⤋
This is not an equation.
If it should be equal to zero, it's fairly easy as
(x-6)^2+(x-6) = (x-6)[(x-6)+1] = (x-6)(x-5)
thus (x-6)(x-5)=0 has two obvious solutions : 5 and 6.
2007-09-18 17:58:38 · answer #3 · answered by JdR 1 · 1⤊ 2⤋
LET X-6=A
SO GIVEN A^2-A
=A(A+1)
=(X-6)(X-6+1) (PUTTING BACK VALUE OF X)
=(X-6)(X-5)
2007-09-18 20:46:30 · answer #4 · answered by Sumita T 3 · 0⤊ 1⤋
Let q = (x-6) and then
q² + q = 0 => q(q + 1) = 0 so
q = 0 or
q = -1
But q = (x-6) so
x - 6 = 0 or
x - 6 = -1 so that
x = 6 or
x = 5
Doug
2007-09-18 17:58:42 · answer #5 · answered by doug_donaghue 7 · 1⤊ 2⤋
Simplify:
= (x - 6)^2 + (x - 6)
= x^2 - 6x - 6x + 36 + x - 6
= x^2 - 11x + 30
Answer: x^2 - 11x + 30
2007-09-18 18:00:25 · answer #6 · answered by Jun Agruda 7 · 4⤊ 1⤋
(x-6)^2 + (x-6) = 0
x^2 - 12 x + 36 + x - 6 = 0
x^2 - 11 x + 30 = 0
(x-6) (x-5) = 0
x= 6
x= 5
2007-09-18 18:01:51 · answer #7 · answered by UJ 3 · 1⤊ 2⤋
(x-6)^2+(x-6)
(x-6) [ (x-6) + 1 ]
(x-6) (x-5)
2007-09-18 17:56:32 · answer #8 · answered by Pakyuol 7 · 1⤊ 2⤋
this is basically
= (x-6)(x-6) + (x-6)
= (x^2 -12x + 36) + (x-6)
= (x^2 - 12x +x +36 - 6)
= x^2 -11x +30
= (x-6)(x-5)
2007-09-18 17:58:50 · answer #9 · answered by princess 3 · 1⤊ 1⤋
ANs.
(x-6)^2+(x-6)=2x-12+x-6
3x=18
x=6.
2007-09-18 18:01:22 · answer #10 · answered by Sasi Kumar 4 · 0⤊ 3⤋