7^x + 4*7^x = 245
5*7^x = 245
7^x = 49
since 49 = 7^2, x=2
x-√x+20 = 0
think of this as
(√x)^2 - √x + 20 = 0
I think you must have a typo. It is much better if you have
(√x)^2 - √x - 20 = 0
(√x-5)(√x+4) = 0
√x = 5 or √x = -4
since √x is always positive, the only solution is
√x = 5
x = 25
1 + 3/(x+1) = x
(x+1) + 3 = x(x+1)
x^2-4 = 0
(x-2)(x+2) = 0
x = -2 or 2
Mismatched parentheses. I will assume you meant
(√24 + 2√6 + √54)/(√96 - √6)
If you factor out the perfect squares, you have
(2√6 + 2√6 + 3√6)/(4√6 - √6)
7√6 / 3√6
7/3
Please help solve for x
A.7^x+(4)(7^x) =245
B.x-|x+20=0
C.1+3/(x+1)=x
Simplify
D.(|24 +2|6 + |54 / |96 - |6
Keys: |=square root, /over or divided by
2 answers
Thank you very much steve god bless you