Question

The sum of roots of the equation $(1.25{)}^{1-x^2}=(0.4096{)}^{1+x}$

• A. Infinite
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

${\left(1.25\right)}^{1-x^2}={\left(0.4096\right)}^{1+x}$

${\left(\frac{125}{100}\right)}^{1-x^2}={\left(\frac{4096}{10000}\right)}^{1+x}$

${\left(\frac{5}{4}\right)}^{1-x^2}={\left({\left(\frac{8}{10}\right)}^4\right)}^{1+x}$

${\left(\frac{5}{4}\right)}^{1-x^2}={\left(\frac{4}{5}\right)}^{4+4x}$

${\left(\frac{5}{4}\right)}^{1-x^2}={\left(\frac{5}{4}\right)}^{-4-4x}$

$\Rightarrow 1-x^2=-4-4x$

$x^2-4x-5=0$

$x^2-5x+x-5=0$

$x(x-5)+1(x-5)=0$

$(x+1)(x-5)=0$

$x=-1,5$

Therefore, Sum of the roots of equation is $4$