Question

Find the Zeroes of the quadratic polynomial $\left(6x^2-7x-3\right)$ and verify the relation between its Zeroes and Coefficients.

Answer 1

Solution:

Step 1: Solve the expression by splitting the middle term:

Given polynomial: $6x^2-7x-3$

$$\begin{gathered} 6x^2-7x-3 \\ =6x^2+2x-9x-3\left(\because -7x=2x-9x\right) \\ =2x(3x+1)-3(3x+1) \\ =(3x+1)(2x-3)....\left(1\right)(\mathrm{taking}\mathrm{out}\mathrm{the}\mathrm{common}\mathrm{factor}) \end{gathered}$$

Step 2: Find zeroes of the expression (1):

$(3x+1)(2x-3)=0(\mathrm{equating}\mathrm{expression}(1)\mathrm{to}\mathrm{zero})$

$$\begin{gathered} \Rightarrow (3\mathrm{x}+1)=0\mathrm{and}(2\mathrm{x}-3)=0 \\ \Rightarrow \mathrm{x}=\frac{\mathbf{-}\mathbf{1}}{\mathbf{3}}\mathrm{and}\mathrm{x}=\frac{\mathbf{3}}{\mathbf{2}} \end{gathered}$$

Step 3: Verify relation between the Zeroes and Coefficients:

$$\begin{gathered} \mathrm{Sum}\mathrm{of}z\mathrm{eroes}=\frac{-(\mathrm{Coefficient}\mathrm{of}\mathrm{x})}{(\mathrm{Coefficient}\mathrm{of}{\mathrm{x}}^2)} \\ =\frac{-1}{3}+\frac{3}{2} \\ =\frac{7}{6}(\because \mathrm{LCM}(2,3)=6) \\ \mathrm{Product}\mathrm{of}z\mathrm{eroes}=\frac{(\mathrm{Constant}\mathrm{term})}{(\mathrm{Coefficient}\mathrm{of}\mathrm{x})} \\ =\frac{-1}{3}\times \frac{3}{2} \\ =\frac{-3}{6} \end{gathered}$$

Final answer: Hence, the zeroes of the polynomial $6x^2-7x-3\mathrm{are}\frac{\mathbf{-}\mathbf{1}}{\mathbf{3}}\mathbf{}\mathbf{,}\frac{\mathbf{3}}{\mathbf{2}}$, it's zeroes and coefficients are verified.