Question

If alpha and beta are the zeros of quadratic polynomial f x power 6 X square + X - 2 find the value of Alpha upon beta and beta upon Alpha

Answer 1

Step-by-step explanation:

Correct Question :-

If α and β are zeroes of polynomial f(x) = 6x² + x - 2. Find the value of α/β + β/α

Solution :-

Given -

• α and β are zeroes of polynomial f(x) = 6x² + x - 2

To Find -

• Value of α/β + β/α

→ α² + β²/αβ

Now,

→ 6x² + x - 2

As we know that :-

• αβ = c/a

→ αβ= -2/6

And

• α + β = -b/a

→ α + β = -1/6

Squaring both sides :-

→ (α + β)² = (-1/6)²

→ α² + β² + 2αβ = 1/36

→ α² + β² = 1/36 - 2×-2/6

→ α² + β² = 1/36 + 4/6

→ α² + β² = 1 + 24/36

→ α² + β² = 25/36

Now,

The value of α² + β²/αβ is

→ 25/36 ÷ -1/6

→ 25/36 × -6

→ -25/6

Hence,

The value of α/β + β/α is -25/6.

Answer 2

Given :

The polynomial is 6(x)² + x - 2 and their roots are α and β

We know that ,

$\large \sf \fbox{ \alpha \beta = \frac{c}{a} }$

Thus ,

$\sf \hookrightarrow \alpha \beta = - \frac{ 2}{6} \\ \\ \sf \hookrightarrow \alpha \beta = - \frac{1}{3}$

And

$\sf \large \fbox{ \alpha + \beta = - \frac{b}{a} }$

Thus ,

$\alpha + \beta = - \frac{1}{6}$

Squaring both sides , we get

$\sf \hookrightarrow { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = - \frac{1}{36} \\ \\\sf \hookrightarrow { \alpha }^{2} + { \beta }^{2} +2 ( - \frac{1}{3} ) = - \frac{1}{36} \\ \\ \sf \hookrightarrow { \alpha }^{2} + { \beta }^{2} = - \frac{1}{36} + \frac{2}{3} \\ \\ \sf \hookrightarrow { \alpha }^{2} + { \beta }^{2} = \frac{ - 1 + 24}{36} \\ \\ \sf \hookrightarrow { \alpha }^{2} + { \beta }^{2} = \frac{ 25}{36}$

Now , the value of α² + β²/αβ will be

$\sf \hookrightarrow \frac{25}{36 \times ( - \frac{1}{3}) } \\ \\\sf \hookrightarrow - \frac{25}{12}$

$\therefore \sf \underline{The \: required \: value \: is \: - \frac{25}{12} }$