Question

if alpha and beta are the zeros of the polynomial 6x2+x-2 find alpha^2/beta+ beta^2/alpha​

Answer 1

Given :-

6x² + x - 2

To Find :-

α²/β + β²/α

Solution :-

6x² + (4x - 3x) + 2 = 0

6x² + 4x - 3x + 2 = 0

2x(3x + 2) - 1(3x + 2) = 0

(2x - 1)(3x + 2) = 0

Either

2x - 1 = 0

2x = 1

x = 1/2

or

3x + 2 = 0

3x = -2

x = -2/3

Now

Finding α²/β + β²/α

(1/2)²/(-2/3) + (-2/3)²/(1/2)

(1/4)/(2/3) + (4/9)/(1/2)

1/4 × 3/2 + 4/9 × 2

3/8 + 8/9

27 + 64/72

91/72

Answer 2

$\sf\large\underline{Given:-}$

$\rm{\longrightarrow P(x)=6x^2+2x-2}$

$\sf\large\underline{To\:Find:-}$

$\rm{\longrightarrow The\:value\:of\dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha}=?}$

$\sf\large\underline{Explanation:-}$

$\rm{\longrightarrow 6x^2+2x-2=0}$

$\rm{\longrightarrow 6x^2-3x+4x-2=0}$

$\rm{\longrightarrow 3x(2x-1)+2(2x-1)=0}$

$\rm{\longrightarrow (3x+2)(2x-1)=0}$

$\rm{\longrightarrow \therefore x=1/2\:,-2/3}$

• $\sf{\alpha=1/2\:\:,\beta=-2/3}$

$\rm{\longrightarrow The\:value\:of\dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha}=?}$

$\rm{\longrightarrow \dfrac{(\frac{1}{2})^2}{\frac{-2}{3}}+\dfrac{(\frac{-2}{3})^2}{\frac{1}{2}}}$