Question

if alpha and beta are the zeros of the polynomial 6x² - 7x + 2 , find a quadratic polynomial whose zeros are 1/alpha and 1/beta​

Answer 1

refer the image..............

Answer 2

Answer:

The required quadratic polynomial is $2x^2+7x+6=0$.

Step-by-step explanation:

Step 1 of 2

It is given that $\alpha$ and $\beta$ are the zeros of the polynomial $6x^2-7x+2$.

Consider the given quadratic equation as follows:

$6x^2-7x+2=0$ . . . . . (1)

Using middle-term splitting method, find the zeros of the polynomial (1) as follows:

⇒ $6x^2-4x-3x+2=0$

⇒ $2x(3x-2)-1(3x-2)=0$

⇒ $(3x-2)(2x-1)=0$

⇒ $x=\frac{2}{3}$ and $x=\frac{1}{2}$

Thus, $\alpha=\frac{2}{3}$ and $\beta=\frac{1}{2}$ are the zeros of the polynomial $6x^2-7x+2$.

Step 2 of 2

Find the quadratic polynomial whose zeros are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

Consider a polynomial whose zeros are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ as follows:

$x^2+(\gamma +\delta)x+\gamma\delta=0$ . . . . . (2)

where $\gamma =\frac{1}{\alpha}$ and $\delta=\frac{1}{\beta}$.

a) Find the value $\gamma$.

Since $\gamma =\frac{1}{\alpha}$

⇒ $\gamma=\frac{1}{2/3}$

⇒ $\gamma=\frac{3}{2}$

b) Find the value $\delta$.

Since $\delta =\frac{1}{\beta}$

⇒ $\delta=\frac{1}{1/2}$

⇒ $\delta=2$

Now,

Substitute the values of $\gamma$ and $\delta$ in the polynomial (2) as follows:

⇒ $x^2+(\frac{3}{2} +2)x+\frac{3}{2} \times2=0$

Simplify as follows:

⇒ $x^2+(\frac{3+4}{2} )x+3=0$

⇒ $x^2+\frac{7}{2}x+3=0$

⇒ $2x^2+7x+6=0$

Therefore, the required quadratic polynomial is $2x^2+7x+6=0$.

#SPJ2