Question

find the zero of the each of the following and verify the relationship between the zeros and their Coefficient 7 x square - 25 x -12​

Answer 1

Given :

find the zero of the each of the following and verify the relationship between the zeros and their Coefficient 7 x² - 25 x -12

To find :

Find the zeros and verify the relationship between zeros and their Coefficient

Solution :

7x² - 25x -12 = 0

solve this by splitting middle term

= 7x² - 3x + 28x - 12 = 0

= x(7x-3) + 4(7x -3) = 0

= (x+4)(7x-3) = 0

Either

=> x + 4 = 0

=> x = -4

or

=> 7x - 3 = 0

=> 7x = 3

=> x = 3/7

-4 and 3/7 are the zeros of the given polynomial

Verification :

$\large{\boxed{\red{\bf{Sum\:of\:zeros}}}}$

Add both the zeros

$\sf -4+\frac{3}{7}=\frac{-25}{7}=\large\frac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}$

$\large{\boxed{\red{\bf{Product\:of\:zeros}}}}$

Multiply both the zeros

$\sf -4\times\frac{3}{7}=\frac{-12}{7}=\large\frac{(constant\:term)}{(coefficient\:of\:x^2)}$

Hence , it is verified

Answer 2

Step-by-step explanation:

Given:

• 7x² – 25x – 12

To Find:

• Zeros of the polynomial and also verify the relationship between zeros and their coefficient.

Solution: Let the given polynomial be denoted by f(x). Then,

$\small\implies{\sf }$ f(x) = 7x² – 25x – 12

$\small\implies{\sf }$ 7x² – 21x – 4x – 12 [ By middle term splitting ]

$\small\implies{\sf }$ 7x (x – 3) – 4 (x – 3)

$\small\implies{\sf }$ (7x – 4) (x – 3)

∴ f(x) = 0 = (7x – 4) (x – 3) = 0

$\small\implies{\sf }$ 7x – 4 = 0 or x – 3 = 0

$\small\implies{\sf }$ x = 4/7 or x = 3/1

So, ★ The zeros of f(x) are 4/7 and 3/1

Sum of zeros = (4/7 + 3) = 4+21/7 = 25/7 = –(coefficient of x) / (coefficient of x²)

★ Product of zeros = 4/7 x 3 = 12/7 = constant term / coefficient of x²