Please show steps and explain.
2006-09-27 13:22:24 · 4 answers · asked by Paul B 1 in Science & Mathematics ➔ Mathematics
2(x - 5)^2 - 7 = 0
2(x - 5)^2 = 7
(x - 5)^2 = (7/2)
x - 5 = sqrt(7/2)
x - 5 = (sqrt(14))/2
x = 5 ± (sqrt(14)/2)
ANS : x = (1/2)(10 ± sqrt(14)) or just 5 ± (sqrt(14)/2), depending on how you want to type it.
2006-09-27 14:48:48 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
Did you copy the question correctly??? Here is what I get.
2(x-5)^2-7=0
2[(x-5)(x-5)]-7=0
2[(x^2-10x+25]-7=0
2x^2-20x+50-7=0
2x^2-20x+43=0
Now the problem from here is that 43 is a prime number and you cant add the factors together to get -20x.
I do know that x does not equal 5. Here is why:
2(5-5)^2-7=0
2(0)^2-7=0
0^2-7=0
0-7 does not equal 0
2006-09-27 21:12:07 · answer #2 · answered by danjlil_43515 4 · 0⤊ 0⤋
2(x-5)^2 - 7 = 0
or, 2(x-5)^2 = 7
or, (x-5)^2 = 7/2
or, x-5 = +/-sqrt(7/2) [ +/- means plus minus ]
or, x = 5 +/- sqrt(7/2)
2006-09-27 22:50:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Solve it yourself and actually learn something.
They should just start calling this site "Homework Cheats"
2006-09-27 20:29:43 · answer #4 · answered by ? 5 · 0⤊ 0⤋