The product of all the real roots of x^2-8x+9 | vedclass

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

Question

(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$ Let $t = x+\frac{1}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$ Let $t = x+\frac{1}{x}$. Then $t^2 = x^2+2+\frac{1}{x^2}$,so $x^2+\frac{1}{x^2} = t^2-2$. Substituting into the equation: $(t^2-2) - 8t + 9 = 0$ $t^2 - 8t + 7 = 0$ $(t-7)(t-1) = 0$ Case $1$: $t = 7$ $\Rightarrow x+\frac{1}{x} = 7$ $\Rightarrow x^2-7x+1 = 0$. The discriminant $D = (-7)^2 - 4(1)(1) = 45 > 0$,so there are two real roots. The product of these roots is $c/a = 1$. Case $2$: $t = 1$ $\Rightarrow x+\frac{1}{x} = 1$ $\Rightarrow x^2-x+1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 Thus,the product of all real roots is $1$.

The product of all the real roots of $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is

Similar Questions

If the roots of the equation $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$ are equal,then in which progression are $a, b, c$?

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

$\alpha, \beta, \gamma, 2, \varepsilon$ are the roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. If $\alpha < \beta < \gamma < 2 < \varepsilon$,then find the value of $\alpha+2 \beta+3 \gamma+5 \varepsilon$.

If $a$ and $b$ are natural numbers such that $2013 + a^2 = b^2$,then the minimum possible value of $ab$ is:

For which value of $k$,are the roots of the equation $2x^2 + 5x + k = 0$ rational?

Vedclass Test Series

Exam Paper Generator

Online Exam Module