The roots of the equations 2x^2 - 5x + 1 = 0 | vedclass

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

Question

(B) Let the roots of the first equation $2x^2 - 5x + 1 = 0$ be $\alpha$ and $\beta$.Then,$\alpha \beta = \frac{c}{a} = \frac{1}{2}$.Let the roots of t

Vedclass · Accepted Answer

Reciprocal and of opposite sign

Vedclass · Answer

(B) Let the roots of the first equation $2x^2 - 5x + 1 = 0$ be $\alpha$ and $\beta$.Then,$\alpha \beta = \frac{c}{a} = \frac{1}{2}$.Let the roots of the second equation $x^2 + 5x + 2 = 0$ be $\gamma$ and $\delta$.Then,$\gamma \delta = \frac{2}{1} = 2$.Comparing the two equations,we observe that the second equation is of the form $cx^2 + bx + a = 0$ where $a=2, b=-5, c=1$ is transformed into $ax^2 - bx + c = 0$ or similar forms.Specifically,if the roots of $ax^2 + bx + c = 0$ are $\alpha, \beta$,then the roots of $cx^2 + bx + a = 0$ are $\frac{1}{\alpha}, \frac{1}{\beta}$.Here,the first equation is $2x^2 - 5x + 1 = 0$ and the second is $x^2 + 5x + 2 = 0$. Note that the coefficients are swapped and the sign of the middle term is changed.If we replace $x$ with $-1/x$ in $2x^2 - 5x + 1 = 0$,we get $2(-1/x)^2 - 5(-1/x) + 1 = 0$,which simplifies to $\frac{2}{x^2} + \frac{5}{x} + 1 = 0$,or $x^2 + 5x + 2 = 0$.Thus,the roots of the second equation are the negative reciprocals of the roots of the first equation. Therefore,they are reciprocal and of opposite sign.

The roots of the equations $2x^2 - 5x + 1 = 0$ and $x^2 + 5x + 2 = 0$ are:

Similar Questions

The value of $p$ for which one root of the equation ${x^2} - 30x + p = 0$ is the square of the other,are

If $x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \dots \infty} } }$,then $x =$

Roots of $2 k x^{2}+5 k x+2=0$ are equal if $k$ is equal to

If $x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$,then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}} = ?$

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{12}{\sqrt{x}} - \frac{23}{\sqrt{x}} = 5\sqrt{x}$
$II.$ $\frac{\sqrt{y}}{12} - \frac{5\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module