Question

find the zero of the quadratic polynomial and verify the relationship with the zero and the COffecant x²+7x+12​

Answer 1

Given :-

x² + 7x + 12

To Find :-

Zeroes

Solution :-

$\sf\dashrightarrow x^{2} + 7x+12$

$\sf\dashrightarrow x^{2} + (4x+3x)+12$

$\sf\dashrightarrow x^{2} + 4x+3x+12$

$\sf\dashrightarrow x(x + 4) + 3(x+4)$

$\sf\dashrightarrow (x+4)(x+3)$

Either

x + 4 = 0

x = 0 - 4

or

x + 3 = 0

x = 0 - 3

x = -3

We have

α = -4

β = -3

a = 1

b = 7

c = 12

Sum of zeroes = α + β = -b/a

-4 + (-3) = -(7)/1

-4 - 3 = -7/1

-7 = -7

Product of zeroes = αβ = c/a

-4 × -3 = 12/1

12 = 12

Hence, verified

[tex][/tex]

Answer 2

Answer:

let f(x)=x²+7x+12

=x²+3x+4x+12

=×(x+3)+4(x+3)

=(x+3)(x+4)

taking f(x)=0

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

x+3=0,x+4=0

x= -3,x =-4

so,-3and-4 are the zeros of given quadratic polynomial

Now,

sum of zeros=-b/a

(-3)+(-4) = -7/1

-7 = -7

product of zeros= c/a

(-3)×(-4)=12/1

12=12

hence ,the relationship between the zeros and coefficient of given polynomial is true