Let x^2=t. Then, 99^t= 100-t. y=99^t is monotonically decreasing and y=100-t is a linear function. Thus, there is, at most, one solution to the given equation. By inspection, it is t=1 > x=+/-1.
Rise both sides to the x^2 power and solve. We'll get x^2=1, so x=1 and x=-1. But x=-1 doesn't match in the original equation. So, only x=1 is a valid solution.
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Let x^2=t. Then, 99^t= 100-t. y=99^t is monotonically decreasing and y=100-t is a linear function. Thus, there is, at most, one solution to the given equation. By inspection, it is t=1 > x=+/-1.
99^t is a function of a^x form, with a=99>1. So it's an increasing function. The rest of the answer is very good(one of 2 likes is mine).
I forgot to mention that y=100-t is decreasing,so we have one increasing and one decreasing functions with only one common solution.
Rise both sides to the x^2 power and solve. We'll get x^2=1, so x=1 and x=-1. But x=-1 doesn't match in the original equation. So, only x=1 is a valid solution.
Why do you say that x = -1 does not match the original equation? I think you are wrong. x=-1 is a valid solution.
@@armacham Sorry, I rushed...x=-1 is a valid solution. I was wrong.
x= -1, +1. Only x^2 appears in the original equation, because (10-x)(10+x) = 100-x^2, so both -1 and +1 are solutions.
X= + -1
χ=-1,ή χ=1
X=1 only
{x^2+x^2 ➖ }=x^4{10x+10x ➖ } =20x^2 (10x)^2=100x^2 {x^4+20x^2}=20x^6 {20x^6 ➖ 100x^2}= 80x^4 8^10x^4 8^2^5x^4 2^3^2^5^1x^4 1^12^1x^2^2 2^1x1^2 1^1x1^2 1x^2 (x ➖ 2x+1) .