Question

Find the zeroes of the polynomial $3{{x}^{2}}-2$ and verify the relationship between the zeroes and coefficients.

Answer 1

Hint: First, using the quadratic formula, find the roots of the given quadratic equation. Then use the relation between the zeroes and the coefficients of a polynomial. Also, try to draw some results seeing the nature of the roots.

Complete step-by-step answer:

The equation given in the question is a quadratic polynomial with the coefficient of x equal to 0.

We know, the roots of the quadratic equation $3{{x}^{2}}-2$ is:

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{0\pm \sqrt{24}}{6}=\pm \sqrt{\dfrac{24}{36}}=\pm \sqrt{\dfrac{2}{3}}$

Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:

Sum of the roots of quadratic equation = $\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}=0$ .

Product of the roots of quadratic equation = \[\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}=\dfrac{-2}{3}\] .

Now using the roots of the mentioned quadratic equation, we get the sum of roots to be equal to $\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{2}{3}}=0$ , and product of roots is equal to $\sqrt{\dfrac{2}{3}}\times \left( -\sqrt{\dfrac{2}{3}} \right)=-\dfrac{2}{3}$ .

Now, as you can see, the results we got using the relationship between the zeroes and coefficients of a quadratic equation and using the actual roots are equal. Therefore, we can say that we have verified the relationship between the zeroes and coefficients for a quadratic polynomial.

Note: If you wanted, you could have directly solved the equation without using the quadratic equation, but if you do so, be careful that ${{x}^{2}}=\dfrac{2}{3}$ implies $x=\pm \sqrt{\dfrac{2}{3}}$ , and not just $x=\sqrt{\dfrac{2}{3}}$.