abc $\neq$ 0 & a, b, c $ϵ$ R. If $x_{1}$ is a root of $a^{2} x^{2} + b x + c = 0, x_{2}$ is a root of $a^{2} x^{2} - b x - c = 0$ and $x_{1} > x_{2} > 0$ , then the equation $a^{2} x^{2} + 2 b x + 2 c = 0$ has a root $x_{3}$ such that

Q. Consider

x^{3} - a x^{2} + b x - c = 0

where

a, b, c

are rationals and

a \neq 1

.It is given that

x_{1}, x_{2}, x_{1} x_{2}

are roots of the above equation and if

b + c = 2 (a + 1)

then

x_{1} x_{2} =

______________.

Join BYJU'S Learning Program

If x1 and x2 are the roots of the equation e2xlnx=x3 with x1>x2, then

The correct option is B x1=x22Clearly, x>0 e2xlnx=x3 ⇒lne2+lnx⋅lnx=3lnx⇒2+(lnx)2=3lnx Let lnx=t Then t2−3t+2=0 ⇒t=1,2∴x1=e2, x2=e⇒x1=x22

If $x_{1}$ and $x_{2}$ are the roots of the equation $e^{2} x^{ln x} = x^{3}$ with $x_{1} > x_{2}$ , then

The correct option is B $x_{1} = x_{2}^{2}$
Clearly, $x > 0$
$e^{2} x^{ln x} = x^{3}$
$\Rightarrow ln e^{2} + ln x \cdot ln x = 3 ln x \Rightarrow 2 + (ln x)^{2} = 3 ln x$
Let $ln x = t$
Then $t^{2} - 3 t + 2 = 0$
$\Rightarrow t = 1, 2 ∴ x_{1} = e^{2}, x_{2} = e \Rightarrow x_{1} = x_{2}^{2}$