Question

find the zeroes of the quadratic polynomial3z²-2z²-1
the
and verify the relation between
zeroes and the coefficients.​

Answer 1

$\bigstar$ Explanation $\red\bigstar$

$\leadsto$ Solution:-

$\longrightarrow$ First, we have find the roots of the quadratic polynomial $3z^2 - 2z - 1$

$3z^2 - 2z -1 =3z^2 - 3z + z - 1 = 3z(z - 1) + 1(z - 1) = (3z+1)(z-1)$ [By splitting the middle term]

$\longrightarrow$ To find zeros of a polynomial, like f(x) then we have to equate f(x) = 0, and we have to find the value of x which is our zero of the polynomial.

Therefore,

Zero of the polynomial   $3z^2 - 2z - 1$ is

$(3z+1)(z-1) = 0$

Here we will get two values of z as it is a quadratic equation

$(3z+1) = 0 \implies z = \frac{-1}{3}$,   $(z-1) = 0 \implies z = 1$

Therefore the zeroes of the polynomial are $\frac{-1}{3}$ and $1$

$\longrightarrow$ Let the coefficient of $z^2$ be a' which is 3

Let the coefficient of $z$ be 'b' which is 2

Let the constant term be 'c' which is 1

Now let's add the two zeroes of the polynomial which we got

Sum of zeroes =  $\frac{-1}{3} +1=\frac{(-1+3)}{3}=\frac{-2}{3}$

The sum of zeroes is equal to $\frac{-b}{a}$

Now let's multiply the zeroes of the polynomial which we got

Product of zeroes = $\frac{-1}{3} \times1$

The product of zeroes is equal to $\frac{-c}{a}$

Therefore,

If the coefficient of $x^2$ be 'a' and the roots are $\alpha$ and $\beta$

Then the quadratic equation will be

$ax^2 - (\alpha +\beta )ax - \alpha \beta =0$.