Question

find the zeros of the quad Polynomials
and verify its relationships between zeros and coefficients 3x²-x-4​

Answer 1

Step-by-step explanation:

here is answer

hope it will be helpful for you!

Answer 2

AnsWer :

x = -1 and x = 4/3.

Solution :

We have equation,

$\tt\leadsto 3 {x}^{2} - x - 4.$

Splitting the middle term.

$\tt \longmapsto 3 {x}^{2} - x - 4 = 0.$

$\tt \longmapsto 3 {x}^{2} + 3x - 4x - 4 = 0.$

$\tt\longmapsto 3x(x + 1) - 4(x + 1) = 0.$

$\tt \longmapsto (3x - 4)(x + 1) = 0.$

Either,

$\tt \mapsto 3x - 4 = 0.$

$\tt\mapsto x = \frac{4}{3}$

Or,

$\tt \mapsto x + 1 = 0.$

$\tt \mapsto x = - 1.$

$\rule{200}3$

Let,

$\tt \mapsto \alpha = - 1. \: \: \: \: \: \beta = \frac{4}{3} .$

Let Verify, It's Zeros.

Sum of Zeros.

$\tt\longmapsto \alpha + \beta = \frac{ - b}{a}= \frac{Coefficient\: of \: x}{Coefficient\: of \: x^2}.$

$\tt\longmapsto \frac{4 - 3}{3} = \frac{1}{3}$

$\tt\longmapsto \frac{1}{3} = \frac{1}{3}$

$\rule{140}1$

Product of Zeros.

$\tt\longmapsto \alpha \beta = \frac{c}{a} = \frac{Constant\: term}{Coefficient \: of \: x^2}$

$\tt\longmapsto \frac{4}{3} . - 1 = \frac{ - 4}{3}$

$\tt\longmapsto \frac{ - 4}{3} = \frac{ - 4}{3}$

Therefore, the zero of given Quadratic polynomial be -1 and 4/3.