Question

9^x^2-1 - 36∗3^x^2-3+3=0

Answer 1

$9^{x^2-1} - 36*3^{x^2-3}+3=0$

$(3^2)^{x^2-1} - 36*3^{x^2-3}+3=0$

$3^{2x^2-2} - 36*3^{x^2-3}+3=0$

$3^{2x^2}*3^{-2} - 36*3^{x^2}*3^{-3}+3=0$

$3^{2x^2}* \frac{1}{9} - 36*3^{x^2}* \frac{1}{27} +3=0$

$\frac{1}{9} *3^{2x^2}- \frac{4}{3} *3^{x^2} +3=0$

$3^{x^2}=a$

$\frac{1}{9} a^2- \frac{4}{3} a +3=0$

$a^2- 12a +27=0$

$D=12^2-4*27=144-108=36$

$a_{1} = \frac{12+6}{2} =9$

$a_{2} = \frac{12-6}{2} =3$

$3^{x^2}=9$

$3^{x^2}=3^2$

$x^{2} =2$

$x= \frac{+}{} \sqrt{2}$


$3^{x^2}=3$

$3^{x^2}=3^1$

$x^{2} =1$

$x= \frac{+}{} 1$