Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $3{x^2} + 5x - 2$, then form a quadratic equation polynomial whose zeroes are $2\alpha$ and $2\beta$.

Answer 1

In the given question, $\alpha$ and $\beta$ are the zeroes of the polynomial $3{x^2} + 5x - 2$. As we know, for a quadratic polynomial $a{x^2} + bx + c$, sum of its zeroes $\alpha$ and $\beta$ is $- \dfrac{b}{a}$ and product of zeroes is $\dfrac{c}{a}$. Using this we will find the value of $\left( {\alpha + \beta } \right)$ and $\alpha \beta$. Also, the quadratic equation in term of roots is given by: ${x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0$. Using this and substituting the values of $\left( {\alpha + \beta } \right)$ and $\alpha \beta$, we will form a quadratic equation polynomial whose zeroes are $2\alpha$ and $2\beta$.

Complete step by step solution:
Given, $\alpha$ and $\beta$ are the zeroes of the polynomial $3{x^2} + 5x - 2$ and we have to form a quadratic equation polynomial whose zeroes are $2\alpha$ and $2\beta$.
As we know, for a quadratic polynomial $a{x^2} + bx + c$, sum of its zeroes $\alpha$ and $\beta$ is $- \dfrac{b}{a}$ and product of zeroes is $\dfrac{c}{a}$.
So, for the given polynomial $3{x^2} + 5x - 2$ with zeroes $\alpha$ and $\beta$, we get
$\Rightarrow \alpha + \beta = - \dfrac{5}{3}$ and $\alpha \beta = - \dfrac{2}{3}$
Now we require a quadratic equation whose zeroes are $2\alpha$ and $2\beta$.
For this polynomial, we get sum of the roots as:
$\Rightarrow 2\alpha + 2\beta = 2\left( {\alpha + \beta } \right)$
$= 2 \times - \dfrac{5}{3}$
On solving, we get
$\Rightarrow 2\alpha + 2\beta = - \dfrac{{10}}{3}$
Similarly, for this polynomial, we get product of the roots as:
$\Rightarrow 2\alpha \times 2\beta = 4\left( {\alpha \beta } \right)$
$= 4 \times \left( { - \dfrac{2}{3}} \right)$
On solving, we get
$\Rightarrow 4\alpha \beta = - \dfrac{8}{3}$
As we know, a quadratic equation is of form:
${x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0$
Therefore, a quadratic equation polynomial whose zeroes are $2\alpha$ and $2\beta$ is given by
$\Rightarrow {x^2} - \left( {2\alpha + 2\beta } \right)x + \left( {4\alpha \beta } \right) = 0$
Putting the values, we get
$\Rightarrow {x^2} + \dfrac{{10}}{3}x - \dfrac{8}{3} = 0$
Multiplying both the sides by $3$, we get
$\Rightarrow 3{x^2} + 10x - 8 = 0$

Note:
Zeroes are not affected by multiplying each term of the polynomial by a constant. A quadratic equation function may have one, two, or zero zeroes. Zeroes are also called the x-intercept or roots. Also, the y-coordinate of any points lying on the x-axis is zero. So, to find the zeros of a quadratic function, we set $f(x) = 0$.