Question

Find the zeros of the polynomial $3\sqrt{2}{x}^2+13x+6\sqrt{2}$ and verify the relationship between the zeros.

• A. $x=\frac{\sqrt{2}}{3}\text{and}\frac{-3}{\sqrt{2}}$
• B. $x=2\sqrt{2}\text{and}\frac{-3}{\sqrt{2}}$
• C. $x=\frac{-2\sqrt{2}}{3}\text{and}-3\sqrt{2}$
• D. $x=\frac{2\sqrt{2}}{3}\text{and}\frac{-3}{\sqrt{2}}$

Answer 1

Answer: (D)

$x=\frac{2\sqrt{2}}{3}\text{and}\frac{-3}{\sqrt{2}}$

Let us first factorize the given equation $3\sqrt{2}{x}^2+13x+6\sqrt{2}$ as shown below:

$\Rightarrow 3\sqrt{2}{x}^2+13x+6\sqrt{2}=3\sqrt{2}{x}^2+9x+4x+6\sqrt{2}=3x(\sqrt{2}x+3)+2\sqrt{2}(\sqrt{2}x+3)=(3x+2\sqrt{2})(\sqrt{2}x+3)$

Therefore, the zeroes of the given polynomial are:

$(3x+2\sqrt{2})=0\Rightarrow x=-\frac{2\sqrt{2}}{3}=\alpha (\sqrt{2}x+3)=0\Rightarrow x=-\frac{3}{\sqrt{2}}=\beta$

Now, the sum and product of the roots is as follows:

$\alpha +\beta =-\frac{2\sqrt{2}}{3}-\frac{3}{\sqrt{2}}=-(\frac{2\sqrt{2}}{3}+\frac{3}{\sqrt{2}})=\frac{-4-9}{3\sqrt{2}}=-\frac{13}{3\sqrt{2}}=-\frac{b}{a}\alpha \beta =(-\frac{2\sqrt{2}}{3})(-\frac{3}{\sqrt{2}})=2=\frac{6\sqrt{2}}{3\sqrt{2}}=\frac{c}{a}$

Hence verified.