Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and co efficient 3x square -4x-7

Answer 1

Step-by-step explanation:

Given:

Polynomial = 3x² - 4x - 7

To find:

Verifying the relationship between zeros of polynomial and it's co-officients.

Solution:

Find zeros of 3x² - 4x - 7

$\sf{\longrightarrow{{3x}^{2} - 4x -7}}$

$\sf{\longrightarrow{ {3x}^{2} + 3x - 7x - 7}}$

$\sf{\longrightarrow{3x(x - 1) - 7(x - 1)}}$

$\sf{\longrightarrow{(3x - 7)(x + 1) = 0}}$

The zeros :

• (3x - 7)

$\sf{\longrightarrow} \: 3x - 7 = 0$

$\sf{\longrightarrow} \: 3x = 7$

$\sf{\longrightarrow} \: x = \dfrac{7}{3}$

• (x + 1)

$\sf{\longrightarrow} \: x + 1 = 0$

$\sf{\longrightarrow} \: x = - 1$

The zeros are $\sf{\dfrac{7}{3}}$ and (-1).

_____________________

Verifying the relationship:

Sum of zeros -

$\sf{\longrightarrow{ \dfrac{7}{3} + ( - 1) }}$

$\sf{\longrightarrow{ \dfrac{7 - 3}{3}}}$

$\sf{\longrightarrow{ \dfrac{4}{3} }}$

Product of zeros:

$\sf{\longrightarrow{ \dfrac{7}{3} \times ( - 1) }}$

$\sf{\longrightarrow{ \dfrac{ -7 }{3}}}$

For polynomial 3x² - 4x - 7, if zeros are α and β.

In the polynomial -

• a = 3
• b = -4
• c = -7

Sum of zeros :

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

$\sf{\longrightarrow} \: \alpha + \beta = \dfrac{ - ( - 4)}{3}$

$\sf{\longrightarrow} \: \alpha + \beta =\dfrac{4}{3}$

Sum of the zeros is $\sf{\dfrac{ - ( - 4)}{3}}$

_____________________

Product of zeros :

$\sf{\longrightarrow} \: \alpha \times \beta = \dfrac{c}{a}$

$\sf{\longrightarrow} \: \alpha \times \beta = \dfrac{-7}{3}$

Product of zeros = $\sf{\dfrac{-7}{3}}$

Hence verified!

Answer 2

Step-by-step explanation:

To find:-

Zeros of the polynomial $\sf 3x^2-4x-7$

Solution:-

$\\ \qquad\quad\rm{:}\dashrightarrow 3 {x}^{2} - 4x - 7 = 0 \\ \\ \qquad\quad\rm{:}\dashrightarrow 3 {x}^{2} + 3x - 7x - 7 = 0 \\ \\ \qquad\quad\rm{:}\dashrightarrow 3x(x + 1) - 7(x + 1) = 0 \\ \\ \qquad\quad\rm{:}\dashrightarrow (x + 1) (3x - 7) = 0 \\ \\ \qquad\quad\rm{:}\dashrightarrow (x + 1) = 0 \: or \: (3x - 7) = 0 \\ \\ \qquad\quad\rm{:}\dashrightarrow x = - 1 \: or \: x = \dfrac{7}{3}$

${\mathcal{\therefore Zeros\;of\:the\; polynomial\:3x^2-4x-7\:are\:-1\:and\:\dfrac{7}{3}}}$