Solve the following equation using the quadra | vedclass

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

Question

(D) Given equation: $x + \frac{1}{x} = 3$.Multiply the entire equation by $x$ to clear the denominator: $x^2 + 1 = 3x$.Rearrange into the standard qua

Vedclass · Accepted Answer

$\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}$

Vedclass · Answer

(D) Given equation: $x + \frac{1}{x} = 3$.Multiply the entire equation by $x$ to clear the denominator: $x^2 + 1 = 3x$.Rearrange into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 - 3x + 1 = 0$.Here,$a = 1$,$b = -3$,and $c = 1$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Calculate the discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$.Since $D > 0$,real roots exist.Substituting the values into the formula: $x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.Thus,the solutions are $x = \frac{3+\sqrt{5}}{2}$ and $x = \frac{3-\sqrt{5}}{2}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $x + \frac{1}{x} = 3, x \neq 0$.

Similar Questions

Does $(x-1)^{2}+2(x+1)=0$ have a real root? Justify your answer.

Solve the following quadratic equation using the quadratic formula: $3x^{2} + 2\sqrt{5}x - 5 = 0$.

Find the roots of the following quadratic equations by the factorisation method:
$\frac{2}{5} x^{2}-x-\frac{3}{5}=0$

The discriminant of the equation $3x^2 - 6x + 4 = 0$ is .............

The roots of a quadratic equation $6 x^{2}-13 x+6=0$ are .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module