one leg of a right triangle id 6; the other leg is (2x-3); the hypotenuse is 9. Find the value of x.

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one leg of a right triangle id 6; the other leg is (2x-3); the hypotenuse is 9. Find the value of x.

Mathematics
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a^2 + b^2 = c^2 6^2 + (2x - 3)^2 = 9^2
We got that...but then can't solve.
36 + (2x - 3)(2x - 3) = 81 you need to foil the 2x-3's 36 +4x^2 - 6x - 6x + 9 = 81 add like terms 4x^2 - 12x + 45 = 81 set it = to 0 4x^2 - 12x - 36 = 0 do you know how to solve from here?

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We got to where you are...then can't get anything to add to -12x?
First of all you have a common factor of 4 you should take out... Also, if you look at the sign in front of the 36 that is what tells you if you should add or subtract.. so in this case you are finding factors that subtract to get 12... BUT 4(x^2 - 3x - 9) and you are correct there are no factors of 9 that subtract to get 3. Do you know the quadratic formula yet?
next chapter in the book...but I guess the teacher thinks we should know it.
No, I would guess that there is a problem with one of the numbers either you wrote above or the problem is wrong.. I would think it would come out to be a nice factored number if this is as far as you have gotten, otherwise you are going to get a number with square roots in it....
so what would the sq root look like?
3 + sqrt(9 - 4(1)(-9)) all over 2 3 + sqrt(45) ---------- 2

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