Question

Find the zeroes of following quadratic polynomials and verify the relationship between the zeroes and the co-effecients ` 4x^(2)-x-5​

Answer 1

Given :-

4x² - x - 5

To Find :-

Zeroes of following quadratic polynomials and verify the relationship between the zeroes and the co-effecient

Solution :-

4x² - x - 5

By discrimination method

x = -b ± √(b² - 4ac)/2a

x = -1 ± √[1 - (4)(4)(-5)]/2(4)

x = -1 ± √[1 - (-80)]/8

x = -1 ± √[1 + 80]/8

x = -1 ± √[81]/8

x = -1 ± 9/8

Either

x = 1

or

x = -5/4

Verification

α + β = -b/a

1 + -5/4= -(1)/4

4 - 5/4 = -1/4

-1/4 = -1/4

Now,

αβ = c/a

1 × -5/4 = -5/4

-5/4 = -5/4

Answer 2

Given :

• A quadratic polynomial 4x² - x - 5

To Find :

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

Solution :

$\longmapsto\tt\bf{{4x}^{2}-x-5=0}$

By Splitting Middle Term :

$\longmapsto\tt{{4x}^{2}-(5x-4x)+5=0}$

$\longmapsto\tt{{4x}^{2}-5x+4x+5=0}$

$\longmapsto\tt{x(4x-5)+1(4x-5)=0}$

$\longmapsto\tt{(4x-5)\:\:(x+32)=0}$

• x = 5/4
• x = -1

So , 5/4 and -1 are the zeroes of Quadratic Polynomial 4x²-x-5 .

Here :

• a = 4
• b = -1
• c = -5

Sum of Zeroes :

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

$\longmapsto\tt{\dfrac{5}{4}+(-1)=\dfrac{-(-1)}{4}}$

$\longmapsto\tt{\dfrac{5-4}{4}=\dfrac{1}{4}}$

$\longmapsto\tt\bf{\dfrac{1}{4}=\dfrac{1}{4}}$

Product of Zeroes :

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

$\longmapsto\tt{\dfrac{5}{4}\times{-1}=\dfrac{-5}{4}}$

$\longmapsto\tt\bf{\dfrac{-5}{4}=\dfrac{-5}{4}}$

HENCE VERIFIED