Question

If one root of the quadratic equation 2x² + 5x -3 = 0 is 1/2 then its other root is.​

Answer 1

$\large\underline{\sf{Solution-}}$

↝ Let assume that

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {2x}^{2} + 5x - 3 = 0$

↝ As, it is given that, one root of the equation is 1/2.

↝ Let assume that

$\rm :\longmapsto\:\boxed{ \tt{ \: \alpha = \frac{1}{2}}}$

We know,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

So,

$\rm :\longmapsto\: \alpha \beta = \dfrac{ - 3}{2}$

$\rm :\longmapsto\: \dfrac{1}{2} \times \beta = \dfrac{ - 3}{2}$

$\bf\implies \: \beta \: = \: - \: 3$

Alternative Method :-

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: \alpha + \beta = - \: \dfrac{5}{2}$

$\rm :\longmapsto\: \dfrac{1}{2} + \beta = - \: \dfrac{5}{2}$

$\rm :\longmapsto\: \beta = - \: \dfrac{5}{2} - \dfrac{1}{2}$

$\rm :\longmapsto\: \beta = \: \dfrac{ - 5 - 1}{2}$

$\rm :\longmapsto\: \beta = \: \dfrac{ - 6}{2}$

$\bf\implies \: \beta \: = \: - \: 3$

Additional Information :-

1. For cubic equation

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \: a {x}^{3} + b {x}^{2} + cx + d = 0, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$