Question

Find the zeroes of the quadratic polynomial 4x2 – 6 – 8x and verify the relationship between the zeroes and the coefficients of the polynomial.​

Answer 1

The roots are $x=\frac{2+\sqrt{10}}{2},\frac{2-\sqrt{10}}{2}$.

Step-by-step explanation:

Given : Quadratic polynomial $4x^2-6-8x$

To find : The zero of the quadratic polynomial and verify the relationship between the zeroes and the coefficients of the polynomial ?

Solution :

Let $\alpha,\beta$ are the roots of the equation.

The polynomial in equation form, $4x^2-8x-6=0$

$2x^2-4x-3=0$

Here, a=2, b=-4 and c=-3.

Using quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(2)(-3)}}{2(2)}$

$x=\frac{4\pm\sqrt{40}}{4}$

$x=\frac{4\pm2\sqrt{10}}{4}$

$x=\frac{2\pm\sqrt{10}}{2}$

The roots are $\alpha=\frac{2+\sqrt{10}}{2},\beta=\frac{2-\sqrt{10}}{2}$

The sum of roots $\alpha+\beta=-\frac{b}{a}$

$\frac{2+\sqrt{10}}{2}+\frac{2-\sqrt{10}}{2}=-\frac{-4}{2}$

$\frac{2+\sqrt{10}+2-\sqrt{10}}{2}=2$

$\frac{4}{2}=2$

$2=2$

Verified.

The product of roots $\alpha\beta=\frac{c}{a}$

$(\frac{2+\sqrt{10}}{2})(\frac{2-\sqrt{10}}{2})=\frac{-3}{2}$

$\frac{2^2-\sqrt{10}^2}{4}=-\frac{3}{2$

$\frac{4-10}{4}=-\frac{3}{2$

$\frac{-6}{4}=-\frac{3}{2$

$-\frac{3}{2}=-\frac{3}{2$

Verified.

#Learn more

In the quadratic equation ax2 + bx + c = 0, if the sum of the roots is equal to the product of the roots, find the sum of the reciprocals of the roots.

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