Question

find the zeroes of the quadratic polynomial y²+92y+1920 and verify the relationship the zeros and the coefficients​

Answer 1

Given :

• A quadratic polynomial y² + 92y + 1920 .

To Find :

• Zeroes of the polynomial and verify the relationship between the zeroes and the coefficients .

Solution :

$\longmapsto\tt\bf{{y}^{2}+92y+1920=0}$

By Splitting Middle Term :

$\longmapsto\tt{{y}^{2}+(60y+32y)+1920=0}$

$\longmapsto\tt{{y}^{2}+60y+32y+1920=0}$

$\longmapsto\tt{y(y+60)+32(y+60)=0}$

$\longmapsto\tt{(y+60)\:\:(y+32)=0}$

• y = -60
• y = -32

So , -60 and -32 are the zeroes of Quadratic Polynomial y² + 92y + 1920 .

.

Here :

• a = 1
• b = 92
• c = 1920

Sum of Zeroes :

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

$\longmapsto\tt{-60+(-32)=\dfrac{-92}{1}}$

$\longmapsto\tt{-92=-92}$

Product of Zeroes :

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

$\longmapsto\tt{-60\times{-32}=\dfrac{1920}{1}}$

$\longmapsto\tt\bf{1920=1920}$

HENCE VERIFIED

Answer 2

Given :-

y² + 92y + 1920

To Find :-

Zeroes

Solution :-

y² + 92y + 1920 = 0

y² + (60y + 32y) + 1920 = 0

y² + 60y + 32y + 1920 = 0

y(y + 60) + 32(y +60) = 0

(y + 60)(y + 32) = 0

So,

Either

y + 60 = 0

y = 0 - 60

y = -60

or

y + 32 = 0

y = 0 - 32

y = 32

Verification

On comparing the given equation with ax² + bx + c. We get,

a = 1

b = 92

c = 1920

Sum of zeroes = α + β = -b/a

-60 + (-32) = -(92)/1

-60 - 32 = -92/1

-92 = -92

Product of zeroes = αβ = c/a

-60 × -32 = 1920/1

1920 = 1920