Question

x^2-1.5x+10=0

Answer 1

yea

Answer 2

wait my bad i dont think its quadratic. i have to use the pythagorean theorem to find all the sides on the triangle. the short side =x the long side=1/2x+11 and the hypotenuse=2x+1

Answer 3

@jdoe0001

Answer 4

well, what I have so far is \(\bf c^2 = a^2 + b^2\\
\color{green}{a = x\\
b = \cfrac{1}{2}x + 11\\
c = 2x + 1}\\
(2x + 1)^2 = (x)^2 + \left(\cfrac{1}{2}x + 11\right)^2\)

Answer 5

how would i solve that

Answer 6

\(\bf (2x + 1)^2 = (x)^2 + \left(\cfrac{1}{2}x + 11\right)^2
\implies (2x + 1)^2 = (x)^2 + \cfrac{(x+22)^2}{2^2}\)

Answer 7

well, first off, you expand the binomials

Answer 8

\(\bf 2x^2+4x+1 = x^2 + \cfrac{x^2+44x+484}{4}\\
\textit{now let's multiply by 4 each side}\\
4(2x^2+4x+1) = \cancel{4}\left(x^2 + \cfrac{x^2+44x+484}{\cancel{4}}\right)\)

Answer 9

so we multiply by 4 just to get rid of the denominator

Answer 10

so that'd leave us with \(\bf 8x^2+16x+4 = 4x^2 + x^2+44x+484\)
when add up any like terms

Answer 11

so, that will give you a quadratic equation, which will give you 2 values for "x"
one is negative btw, so we can drop that one off as non-valid and keep the positive one only