Sovle
w^2-13w+42=0
I get -6,-7 is this correct????
yes it is, as:
w^2-13w+42=0
(w-7)(w-6)=0
You correctly factored the trinomial, but that's not the solution.
If (x-6)(x-7) = 0, then the equation is true only if one of the factors is zero.
So, the solution is, x=6 or 7
To solve the quadratic equation w^2 - 13w + 42 = 0, we can use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -13, and c = 42. Plugging the values into the formula, we get:
w = (-(-13) ± √((-13)^2 - 4(1)(42))) / (2(1))
w = (13 ± √(169 - 168)) / 2
w = (13 ± √1) / 2
w = (13 ± 1) / 2
So the two possible solutions for w are:
w = (13 + 1) / 2 = 14 / 2 = 7
w = (13 - 1) / 2 = 12 / 2 = 6
Hence, the correct solutions to the equation w^2 - 13w + 42 = 0 are w = 7 and w = 6. Therefore, your answer of -6 and -7 is incorrect.