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Find the zeroes of the quadratic polynomial 2 x^{2}-11 x+15 and verify the relation between the zeroes and the coefficients.
CBSE | Class 10 | Excercise 12A | Maths | Polynomials | RS Aggarwal
Find the zeroes of the quadratic polynomial 2 x^{2}-11 x+15 and verify the relation between the zeroes and the coefficients.

    \[\begin{aligned} \mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}-11 \mathrm{x}+15 \\ &=2 \mathrm{x}^{2}-(6 \mathrm{x}+5 \mathrm{x})+15 \\ &=2 \mathrm{x}^{2}-6 \mathrm{x}-5 \mathrm{x}+15 \\ =& 2 \mathrm{x}(\mathrm{x}-3)-5(\mathrm{x}-3) \\ \quad=(2 \mathrm{x}-5)(\mathrm{x}-3) \\ \therefore \mathrm{f}(\mathrm{x}) &=0 \Rightarrow(2 \mathrm{x}-5)(\mathrm{x}-3)=0 \\ & \Rightarrow 2 \mathrm{x}-5=0 \text { or } \mathrm{x}-3=0 \\ & \Rightarrow \mathrm{x}=\frac{5}{2} \text { or } \mathrm{x}=3 \end{aligned}\]

So, the zeroes of f(x) are \frac{5}{2} and 3 .

Sum of zeroes =\frac{5}{2}+3=\frac{5+6}{2}=\frac{11}{2}=\frac{-(\text { coef ficient of } x)}{\left(\text { coefficient of } x^{2}\right)}

Product of zeroes =\frac{5}{2} \times 3=\frac{-15}{2}=\frac{\text { constant term }}{\left(\text { coef ficient of } x^{2}\right)}

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