Question

ifα and β are the zeroes of the quadratic polynomial f(x)=6x^2+x-2,find the value of α/β+β/α​

Answer 1

Answer:

Step-by-step explanation:

Let’s first solve the equation to get our zeroes of the equation i.e. alpha and beta.

6x2+x−2=0

6x2+4x−3x−2=0

2x(3x+2)−(3x+2)=0

which gives us two equations,

2x−1=0=>x=1/2(alpha)

3x+2=0=>x=−2/3(beta)

So, the value of alpha/beta and beta/alpha will be -3/4 and -4/3, adding both of them gives the solution as -25/12.

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-25}{12}}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies \alpha \: and \: \beta \in( 6 {x}^{2} + x - 2 = 0) \\ \\ \underline \bold{to \: find: } \\ \implies \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = ?$

• According to given question :

$\bold{using \: quadratic \: formula : } \\ \implies 6{x}^{2} + x - 2 = 0 \\ \\ \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \implies x = \frac{ - 1 \pm \sqrt{ {1}^{2} - 4 \times 6 \times ( - 2)} }{2 \times 6} \\ \\ \implies x = \frac{ - 1 \pm \sqrt{1 + 48} }{12} \\ \\ \implies x = \frac{ - 1 \pm \sqrt{49} }{12} \\ \\ \implies x = \frac{ - 1 \pm7 }{12} \\ \\ \bold{\implies x = \frac{1}{2} \: and \: \frac{ - 2}{3} } \\ \\ \bold{\implies \alpha = \frac{1}{2} } \\ \\ \bold{ \implies \beta = \frac{ - 2}{3} } \\ \\ \bold{for \: finding \: value : } \\ \implies \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ \implies \frac{ \frac{1}{2} }{ \frac{ - 2}{3} } + \frac{ \frac{ - 2}{3} }{ \frac{1}{2} } \\ \\ \implies \frac{3}{2 \times - 2} + \frac{ - 2\times 2}{3} \\ \\ \implies \frac{ -3 }{ 4} + \frac{ - 4}{3} \\ \\ \implies \frac{ - 9 + ( - 16)}{12} \\ \\ \implies \frac{ -9 - 16}{12} \\ \\ \bold{\implies \frac{ - 25}{12}}$