Question:
If $y = x + \frac{1}{x}, x \ne 0$, then the equation $\left(x^{2}-3x+1\right)\left(x^{2}-5x+1\right)=6x^{2}$ reduces to
If $y = x + \frac{1}{x}, x \ne 0$, then the equation $\left(x^{2}-3x+1\right)\left(x^{2}-5x+1\right)=6x^{2}$ reduces to
Updated On: Apr 8, 2024
- $y^2-8y + 7 = 0$
- $y^2+8y + 7 = 0$
- $y^2-8y - 9 = 0$
- $y^2-8y + 9 = 0$
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The Correct Option is D
Solution and Explanation
Given, $y=x+\frac{1}{x}$
and $\left(x^{2}-3 x+1\right)\left(x^{2}-5 x+1\right)=6 x$
$\Rightarrow\left(x-3+\frac{1}{x}\right)\left(x-5+\frac{1}{x}\right)=6 ($ divide by $x)$
$\Rightarrow (y-3)(y-5)=6$
$\Rightarrow y^{2}-8 y+15=6$
$\Rightarrow y^{2}-8 y+9=0$
and $\left(x^{2}-3 x+1\right)\left(x^{2}-5 x+1\right)=6 x$
$\Rightarrow\left(x-3+\frac{1}{x}\right)\left(x-5+\frac{1}{x}\right)=6 ($ divide by $x)$
$\Rightarrow (y-3)(y-5)=6$
$\Rightarrow y^{2}-8 y+15=6$
$\Rightarrow y^{2}-8 y+9=0$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
