Question

Find the zeroes of the quadratic polynomialx2+ 7x + 12, and verify the relationship between the zeroes and coefficient​

Answer 1

Given :

• A polynomial x²+7x+12.

To Find :

• Zeroes of the given polynomial and verify the relationship between zeroes and coefficient.

Solution :

$\longmapsto\tt\bf{{x}^{2}+7x+12}$

By Splitting Middle Term :

$\longmapsto\tt{{x}^{2}+(4x+3x)+12}$

$\longmapsto\tt{{x}^{2}+4x+3x+12}$

$\longmapsto\tt{x(x+4)+3(x+4)}$

$\longmapsto\tt{(x+3)(x+4)}$

• x = -3
• x = -4

So , -3 and -4 are the zeroes of polynomial x²+7x+12..

_______________________

Here :

• a = 1
• b = 7
• c = 12

Sum of Zeroes :

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{-3+(-4)=\dfrac{-7}{1}}$

$\longmapsto\tt{-3-4=7}$

$\longmapsto\tt\bf{7=7}$

Product of Zeroes :

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{-3\times{-4}=\dfrac{12}{1}}$

$\longmapsto\tt\bf{12=12}$

HENCE VERIFIED

Answer 2

Given :

A polynomial x²+7x+12.

To Find :

Zeroes of the given polynomial and verify the relationship between zeroes and coefficient.

Solution :

\longmapsto\tt\bf{{x}^{2}+7x+12}⟼x

2

+7x+12

By Splitting Middle Term :

\longmapsto\tt{{x}^{2}+(4x+3x)+12}⟼x

2

+(4x+3x)+12

\longmapsto\tt{{x}^{2}+4x+3x+12}⟼x

2

+4x+3x+12

\longmapsto\tt{x(x+4)+3(x+4)}⟼x(x+4)+3(x+4)

\longmapsto\tt{(x+3)(x+4)}⟼(x+3)(x+4)

x = -3

x = -4

So , -3 and -4 are the zeroes of polynomial x²+7x+12..

_______________________

Here :

a = 1

b = 7

c = 12

Sum of Zeroes :

\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}⟼α+β=

a

−b

\longmapsto\tt{-3+(-4)=\dfrac{-7}{1}}⟼−3+(−4)=

1

−7

\longmapsto\tt{-3-4=7}⟼−3−4=7

\longmapsto\tt\bf{7=7}⟼7=7

Product of Zeroes :

\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}⟼αβ=

a

c

\longmapsto\tt{-3\times{-4}=\dfrac{12}{1}}⟼−3×−4=

1

12

\longmapsto\tt\bf{12=12}⟼12=12

HENCE VERIFIED