Question

Find a quadratic polynomial whose zeroes are$-\sqrt{3}$and$\dfrac{-7}{\sqrt{3}}$.

Answer 1

Hint:In this problem, the zeroes of a quadratic polynomial or the roots of a quadratic equation are given. We know if $\alpha $and $\beta $are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$,then $\alpha +\beta =\dfrac{-b}{a}$and $\alpha \beta =\dfrac{c}{a}$.

Complete step-by-step answer:

According to the question, two roots are given therefore it is a quadratic equation.

If $f(x)$is a quadratic polynomial then $f(x)=0$is called a quadratic equation.

Therefore, $\alpha =-\sqrt{3}$and$\beta =\dfrac{-7}{\sqrt{3}}$ are the two roots of the quadratic equation according to the question.

We know, a quadratic equation is represented as $a{{x}^{2}}+bx+c=0$ . According to the hint,

$\begin{align}

& \Rightarrow \alpha +\beta =\dfrac{-b}{a}.......(i) \\

& \Rightarrow \alpha \beta =\dfrac{c}{a}..........(ii) \\

\end{align}$

For the above question equation (i) can be written as,

$\begin{align}

& \Rightarrow \alpha +\beta =-\sqrt{3}+\dfrac{-7}{\sqrt{3}} \\

& \Rightarrow \dfrac{\left( -\sqrt{3}\times \sqrt{3} \right)+-7}{\sqrt{3}}=\dfrac{-3+-7}{\sqrt{3}} \\

& \Rightarrow \dfrac{-10}{\sqrt{3}} \\

& \therefore \alpha +\beta =\dfrac{-10}{\sqrt{3}} \\

\end{align}$

$\Rightarrow \alpha +\beta =\dfrac{-b}{a}=\dfrac{-10}{\sqrt{3}}.......(iii)$

For the above question equation (ii) can be written as,

$\begin{align}

& \Rightarrow \alpha \beta =-\sqrt{{{3}}}\times \dfrac{-7}{\sqrt{{{3}}}} \\

& \therefore \alpha \beta =7 \\

\end{align}$


$\Rightarrow \alpha \beta =\dfrac{c}{a}=7.......(iv)$

We know, a quadratic equation is of the form,

$a{{x}^{2}}+bx+c=0......(v)$

From equation (iii) $b$can be written as,

$\Rightarrow \dfrac{-b}{a}=\dfrac{-10}{\sqrt{3}}$

By cross multiplication the above equation can be written as,

$\begin{align}

& \Rightarrow -b=\dfrac{-10a}{\sqrt{3}} \\

& \Rightarrow b=\dfrac{10a}{\sqrt{3}} \\

& \therefore b=\dfrac{10a}{\sqrt{3}}......(vi) \\

\end{align}$

From equation (iv) $c$can be written as,

$\Rightarrow \dfrac{c}{a}=7$

By cross multiplication, the above equation can be written as,

$\begin{align}

& \Rightarrow c=7a \\

& \therefore c=7a.....(vii) \\

\end{align}$

Substituting equation (vi) and equation (vii) in equation (v) we get as,

$\Rightarrow a{{x}^{2}}+\dfrac{10}{\sqrt{3}}ax+7a=0.....(viii)$

Taking $a$as common in equation (viii), it can be written as,

$\Rightarrow a\left( {{x}^{2}}+\dfrac{10}{\sqrt{3}}x+7 \right)=0.....(ix)$

From equation (ix) we observe that:

Either $a=0$ and $\left( {{x}^{2}}+\dfrac{10}{\sqrt{3}}x+7 \right)\ne 0$ or,

Either $a\ne 0$ and $\left( {{x}^{2}}+\dfrac{10}{\sqrt{3}}x+7 \right)=0$ or,

Both $a=0$ and $\left( {{x}^{2}}+\dfrac{10}{\sqrt{3}}x+7 \right)=0$.

From equation (vi) and equation (vii) we observe that a) and c) are not possible. Hence, the only possible condition is b).

Thus, we can write equation (ix),

$\Rightarrow \left( {{x}^{2}}+\dfrac{10}{\sqrt{3}}x+7 \right)=0$

By taking LCM on RHS of the equation,

\[\Rightarrow (\sqrt{3}{{x}^{2}}+10x+7\sqrt{3})=0......(x)\]

Hence, from equation (x) we know that the quadratic polynomial whose zeroes are$-\sqrt{3}$and$\dfrac{-7}{\sqrt{3}}$ is,

\[\sqrt{3}{{x}^{2}}+10x+7\sqrt{3}\].

Hence, the correct answer is \[\sqrt{3}{{x}^{2}}+10x+7\sqrt{3}\].

Note: The main idea of the problem is that the student should know the relation between the roots or zeroes of a quadratic polynomial or an equation and their coefficients.

This problem can be done in a different way. If $\alpha $and $\beta $are the roots of a quadratic equation, then that quadratic equation can be represented as,
$\Rightarrow {{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0$