Equation frac{3}{{x - {a^3}}} + frac{5}{{x - | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Equation $3\left(x-a^{5}\right)\left(x-a^{7}\right)+5\left(x-a^{3}\right)\left(x-a^{7}\right)+7\left(x-a^{3}\right)\left(x-a^{5}\right)=0$

${\rm{ N

Vedclass · Accepted Answer

Two real and positive roots

Vedclass · Answer

Equation $3\left(x-a^{5} ight)\left(x-a^{7} ight)+5\left(x-a^{3} ight)\left(x-a^{7} ight)+7\left(x-a^{3} ight)\left(x-a^{5} ight)=0$ ${ m{ Now }}\,\,\,a > 1 \Rightarrow {a^3} < {a^5} < {a^7}$ ${ m{ Let }}f(x) = 3\left( {x - {a^5}} ight)\left( {x - {a^7}} ight) + 5\left( {x - {a^5}} ight)\left( {x - {a^7}} ight) + $ ${7\left(x-a^{3} ight)\left(x-a^{5} ight) /\left(x-a^{3} ight)\left(x-a^{3} ight)\left(x-a^{7} ight)} $ $f\left( {{a^{3 - }}} ight) < 0,f\left( {{a^{3 + }}} ight) > 0,f\left( {{a^{5 - }}} ight) < 0$ $f\left( {{a^{5 + }}} ight) > 0,f\left( {{a^{7 - }}} ight) < 0\,\,and\,\,f\left( {{a^{7 + }}} ight) > 0$ $ herefore$ Equation has one root lying in $\left(a^{3}, a^{5} ight)$ and other in $\left(a^{5}, a^{7} ight)$ Now $a>1 herefore$ Roots are real and positive

Equation $\frac{3}{{x - {a^3}}} + \frac{5}{{x - {a^5}}} + \frac{7}{{x - {a^7}}} = 0,a > 1$ has

Similar Questions

If the sum of two of the roots of ${x^3} + p{x^2} + qx + r = 0$ is zero, then $pq =$

The number of real solution(s) of the equation $x^2+3 x+2=\min \{|x-3|,|x+2|\}$ is:

The solution set of the equation $pq{x^2} - {(p + q)^2}x + {(p + q)^2} = 0$ is

If ${x^2} + px + 1$ is a factor of the expression $a{x^3} + bx + c$, then