Question

Find the zeros of quadratic polynomials and verify the relationship between the zeros and their coefficients: f(x) = 6x2 - 3- 7x

Answer 1

GIven,
$$\begin{aligned} &f(x)=6 x^{2}-3-7 x \\ &\text { We put } f(x)=0 \\ &=6 x^{2}-3-7 x=0 \\ &=6 x^{2}-9 x+2 x-3=0 \\ &=3 x(2 x-3)+1(2 x-3)=0 \\ &=(2 x-3)(3 x+1)=0 \end{aligned}$$
This gives us 2 zeros, for
$$x=\frac{3}{2} \text { and } x=\frac{-1}{3}$$
Hence, the zeros of the quadratic equation are $\frac{3}{2}$ and $\frac{-1}{3}$.
Now, for verification Sum of zeros $=\frac{\operatorname{coc} f f i d i e n t ~ o f ~}{\cos f f \text { icicnt of } x^{2}}$
$\frac{3}{2}+\left(\frac{-1}{3}\right)=-\frac{(-7)}{6}$ $\frac{7}{6}=\frac{7}{6}$ Product of roots $=\frac{\text { comstant }}{\text { ceffident of } x^{2}}$ $\frac{3}{2} \times\left(\frac{-1}{3}\right)=\frac{(-3)}{6}$ $\frac{-1}{2}=\frac{-1}{2}$
Therefore, the relationship between zeros and their coefficlents Is verifled.

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