The product of all the rational roots of the | vedclass

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

Question

(A) Given equation: $(x^2-9x+11)^2-(x^2-9x+20)=3$ Let $t = x^2-9x$. Substituting $t$ into the equation: $(t+11)^2 - (t+20) = 3$ $t^2 + 22t + 121 - t -

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given equation: $(x^2-9x+11)^2-(x^2-9x+20)=3$ Let $t = x^2-9x$. Substituting $t$ into the equation: $(t+11)^2 - (t+20) = 3$ $t^2 + 22t + 121 - t - 20 - 3 = 0$ $t^2 + 21t + 98 = 0$ $(t+14)(t+7) = 0$ So,$t = -7$ or $t = -14$. Case $1$: $x^2-9x = -7 \Rightarrow x^2-9x+7 = 0$. The roots are $x = \frac{9 \pm \sqrt{81-28}}{2} = \frac{9 \pm \sqrt{53}}{2}$ (irrational). Case $2$: $x^2-9x = -14 \Rightarrow x^2-9x+14 = 0$. $(x-7)(x-2) = 0 \Rightarrow x = 7, 2$ (rational). The rational roots are $7$ and $2$. The product of the rational roots is $7 	imes 2 = 14$.

The product of all the rational roots of the equation $(x^2-9x+11)^2-(x-4)(x-5)=3$ is equal to:

Similar Questions

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

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2}x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+ax+b=0$. Then the roots of the equation $x^{2}-(a+b-2)x+(a+b+2)=0$ are...

Let $[x]$ be the greatest integer less than or equal to $x$ for a real number $x$. Then the equation $[x^2] = x + 1$ has:

The roots of the equation $x^4 - 2x^3 + x = 380$ are

If two roots of the equation $x^3 - 3x + 2 = 0$ are the same,then the roots are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module