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

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(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

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Reciprocal and of opposite sign

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(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:

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