Question

If a(alpha) and B(beta) are the zeroes of the polynomial 3x^2 + 4x - 4. form a polynomial
whose zeroes are 3a and 3B​

Answer 1

Solution A

Let the new roots be $\alpha '$ and $\beta '$. It is given that the polynomial $3x^2+4x-4$ has $\alpha$ and $\beta$ as zeroes.

The roots of the equation to be formed satisfy,

• $\alpha '=3\alpha$
• $\beta '=3\beta$

Then,

• $\alpha =\dfrac{\alpha '}{3}$
• $\beta =\dfrac{\beta '}{3}$

Substitutions of the zeroes obtain,

• $3(\dfrac{\alpha '}{3} )^2+4(\dfrac{\alpha '}{3} )-4=0\implies \boxed{\alpha '^2+4\alpha '-12=0}$
• $3(\dfrac{\beta '}{3} )^2+4(\dfrac{\beta '}{3} )-4=0\implies \boxed{\beta '^2+4\beta '-12=0}$

The required polynomial is $\boxed{k(x^2+4x-12)}$ where $k$ is a constant.

Solution B

We know that,

$\begin{cases} & \alpha +\beta =-\dfrac{4}{3}\\ & \alpha \beta =-\dfrac{4}{3}\end{cases}$

Then,

$\begin{cases} & 3\alpha +3\beta =-4\\ & 3\alpha \cdot 3\beta =-12\end{cases}$

The new polynomial to be formed is $\boxed{k(x^2+4x-12)}$ where $k$ is a constant.

Answer 2

Required Answer :-

Sum of zeroes = 4

Product of zeroes = 4

Now

$\sf \: 3( \alpha ) + 3( \beta ) = 3( \alpha + \beta ) = 3(4) = 12$

$\sf \: 3 \alpha \times 3 \beta = 3( \alpha \beta ) = 3(4) = 12$

Required Polynomial

$\sf \: {x}^{2} - ( \alpha + \beta ) \times x + ( \alpha \beta )$

$\sf \: {x}^{2} - (12)x - 12$

$\sf \: {x}^{2} - 12x - 12$