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beth12345
how to find the value of 'X' when x^2-8x=8?
\[x^2-8x-8 = 0 \] Use the quadratic formula: \[\frac{-(b)\pm\sqrt{(b)^{2}-4(a)(c)}}{2(a)} \] \[a = 1 , b=-8 , c = -8\] And sub it in.
use the quadratic formula. \[(-b \pm \sqrt{b ^{2} -4ac}) /2a \]
x^2-8x=8 x(x-8)=8 or x=2 and x-8=4 x=2, x=12
thank you, i never thouht of it that way but i can see how it works :)
but this solution is wrong
x=2, x=12 is a wrong solution
Completing the square would work as well. But using the quadratic formula is easier.
i thought it would be like (2)[(12)-8]=8?
oh ok, mimi, i did the formula and got 4+or-2root6, is that right?
Ravi, in case you were wondering why it's wrong it's because you must factorise the equation equal to 0. 8 has an infinite number of non-integer factors, so you can't solve. 0, on the other hand, must have the value of a factor as 0.
thanks samarilli, i see now why it's not the same answer as when i use the formula
Well, i got \[\frac{-8\pm\sqrt{96}}{2} \]