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

If $\alpha$ and $\beta$ are the roots of the equation $2x(2x + 1) = 1$,then $\beta$ is equal to

If the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$ have a common root,then $\left( \frac{a + b}{c} \right)$ is equal to (where $a, b, c \in R$).

Solve the given two equations and choose the correct answer from the given options.
$I.$ $16x^2 + 20x + 6 = 0$
$II.$ $10y^2 + 38y + 24 = 0$

The equation $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$ is a perfect square for how many values of $k$?

Let $p, q, r \in \mathbb{R}$ and $r > p > 0.$ If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta,$ then $|\alpha| + |\beta|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module