ABC is a triangle
1 answer:
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$\implies 2^{2x}-2\cdot5^{2x}-(2^x)(5^x)>0$
let $2^x=a$ and $5^x=b$
So, $a^2-2b^2-ab>0$
divide by $b^2$, ($b^2>0$)
$\implies \left(\frac ab\right)^2-\left(\frac ab\right) -2>0$
this is a quadratic, in $\left(\frac ab\right)$, let it be $x$
So, $x^2-x-2>0$
Can you simplify it now?
Answer:
all work is shown and pictured
