Question

Find zeros of the quadratic polynomial whose sum and product respectively of the zeroes are $-\dfrac{8}{3},\dfrac{4}{3}$.
(a) $-2,\dfrac{1}{3}$
(b) $-2,\dfrac{-2}{3}$
(c) $-2,\dfrac{1}{3}$
(d) $-4,\dfrac{-2}{3}$

Answer 1

Hint: In this question, we will form a polynomial using sum of zeroes and product of zeroes formula and then find zeros using factorization. .

Complete step-by-step answer:
We know that, for a quadratic polynomial of the form $a{{x}^{2}}+bx+c$ with zero $\alpha $ and $\beta $ where $\alpha $ and $\beta $ are real numbers, sum and product of zeroes can be given as follows:
$\alpha +\beta =\dfrac{-b}{a}.........(i)$ and $\alpha \beta =\dfrac{c}{a}.........(ii)$ .
In a given question, let the required zero be $\alpha $ and $\beta $
Then we have,
\[\begin{align}
  & \alpha +\beta =\dfrac{-8}{3} \\
 & and\,\alpha \beta =\dfrac{4}{3} \\
\end{align}\]
Using equation(i) and (ii), we get
\[\begin{align}
  & \dfrac{-8}{3}=\dfrac{-b}{a}......(iii) \\
 & \dfrac{4}{3}=\dfrac{c}{a}........(iv) \\
\end{align}\]
Let us consider \[a=3\]
Then from(iii), we get,
\[\dfrac{-8}{3}=\dfrac{-b}{3}\]
Multiplying \[-3\]both sides, we get,
\[\dfrac{-8}{3}\times \left( -3 \right)=\dfrac{-b}{3}\left( -3 \right)\]
\[\Rightarrow b=8\]
And from equation(iv), we get,
\[\dfrac{4}{3}=\dfrac{c}{3}\]
Multiplying \[3\]on both sides, we get,
 \[\begin{align}
  & \dfrac{4}{3}\times 3=\dfrac{c}{3}\times 3 \\
 & \Rightarrow c=4 \\
\end{align}\]
Therefore, the required quadratic polynomial will be,
\[k\left( 3{{x}^{2}}+8x+4 \right)\], where \[k\]is any scalar number and \[k\ne 0\]
Now, to find zeros of this polynomial, we write,
\[\begin{align}
  & k\left( 3{{x}^{2}}+8x+4 \right)=0 \\
 & \Rightarrow 3{{x}^{2}}+8x+4=0 \\
\end{align}\]
Splitting middle term, we get,
$3{{x}^{2}}+6x+2x+4=0$
Taking common terms, we get,
$3x\left( x+2 \right)+2\left( x+2 \right)=0$
Taking $\left( x+2 \right)$ common, we get,
$\left( x+2 \right)\left( 3x+2 \right)=0$
Therefore, either,or $3x+2=0$
$\Rightarrow $$x=-2$ or $3x=-2$
$\Rightarrow $$x=-2$ or $x=\dfrac{-2}{3}$
Hence the required zeroes are $-2$ and $\dfrac{-2}{3}$.
Therefore, the correct option is (b).

Note: Alternative way: We have $\alpha +\beta =\dfrac{-8}{3}$ .
\[\Rightarrow \alpha =\dfrac{-8}{3}-\beta \]
We also have $\alpha \beta =\dfrac{4}{3}$ .
$\Rightarrow \left( \dfrac{-8}{3}-\beta \right)\beta =\dfrac{4}{3}$
 $\Rightarrow \dfrac{-8\beta }{3}-{{\beta }^{2}}-\dfrac{4}{3}=0$
$\Rightarrow -8\beta -3{{\beta }^{2}}-4=0$
So, the required equation will be $3{{x}^{2}}+8x+4=0$ .
Splitting middle term, we get,
$3{{x}^{2}}+6x+2x+4=0$
Taking common terms, we get,
$3x\left( x+2 \right)+2\left( x+2 \right)=0$
Taking $\left( x+2 \right)$ common, we get,
$\left( x+2 \right)\left( 3x+2 \right)=0$
Therefore, either,or $3x+2=0$
$\Rightarrow x=-2$ or $3x=-2$
$\Rightarrow $$x=-2$ or $x=\dfrac{-2}{3}$
Hence the required zeroes are $-2$ and $\dfrac{-2}{3}$.