Question

Find the zeroes of the Quadratic Polynomial and then, verify the relationship between the zeroes and the coefficients.
$\mathtt{6x^2-3-7x}$

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given Polynomial:

$\rm\longrightarrow f(x)=6x^{2}-7x-3$

$\rm\longrightarrow 6x^{2}-7x-3=0$

$\rm\longrightarrow 6x^{2}-9x+2x-3=0$

$\rm\longrightarrow 3x(2x-3)+1(2x-3)=0$

$\rm\longrightarrow (3x+1)(2x-3)=0$

Therefore:

$\rm\longrightarrow x=\dfrac{-1}{3},\dfrac{3}{2}$

Verification:

Comparing f(x) with ax² + bx + c, we get:

$\rm\longrightarrow\begin{cases}\rm a=6\\ \rm b=-7\\ \rm c=-3\end{cases}$

We know that:

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{-b}{a}$

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{-(-7)}{6}$

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{7}{6}$

Also:

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{-1}{3}+\dfrac{3}{2}$

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{-2+9}{6}$

$\rm\longrightarrow Sum\ Of\ Roots=\dfrac{7}{6}$

Again:

$\rm\longrightarrow Product\ Of\ Roots=\dfrac{c}{a}$

$\rm\longrightarrow Product\ Of\ Roots=\dfrac{-3}{6}$

$\rm\longrightarrow Product\ Of\ Roots=\dfrac{-1}{2}$

Also:

$\rm\longrightarrow Product\ Of\ Roots=\dfrac{-1}{3}\times\dfrac{3}{2}$

$\rm\longrightarrow Product\ Of\ Roots=\dfrac{-1}{2}$

Hence Verified..!!

$\textsf{\large{\underline{Learn More}:}}$

1. Relationship between zeros and coefficients (Quadratic Polynomial)

Let f(x) = ax² + bx + c and let α and β be the zeros of f(x).

Therefore:

$\rm\longrightarrow\alpha+\beta=\dfrac{-b}{a}$

$\rm\longrightarrow\alpha\beta=\dfrac{c}{a}$

2. Relationship between zeros and coefficients (Cubic Polynomial)

Let f(x) = ax³ + bx² + cx + d and let α, β and γ be the zeros of f(x).

Therefore:

$\rm\longrightarrow \alpha+\beta+\gamma=\dfrac{-b}{a}$

$\rm\longrightarrow \alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}$

$\rm\longrightarrow \alpha\beta\gamma=\dfrac{-d}{a}$