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Answer 1

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 :

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

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

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

$\longmapsto\tt{(4x-5)\:\:(x+1=0}</p><p>$

• 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 :

$</p><p>\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

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

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

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

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

Answer 2

$\red{\bigstar}$ G I V E N

$\:$

• Polynomial, P(x) = 4x² - x - 5.

$\:$

$\orange{\bigstar}$ T OㅤF I N D

$\:$

• Zeroes of the given polynomial (4x² - x - 5), also we have to verify relationship between it's zeroes and the coefficients.

$\:$

$\blue{\bigstar}$ S O L U T I O N

$\:$

Finding zeroes of given polynomial (α and β) ::

$\\ \dashrightarrow\:\sf 4x^2 + 4x - 5x - 5 = 0$

$\\ \dashrightarrow\:\sf 4x\big(x + 1\big) - 5\big(x + 1\big) = 0$

$\\ \dashrightarrow\:\sf \big (4x - 5\big) \big(x + 1\big) = 0$

$\\ \dashrightarrow\:\sf 4x - 5 = 0\:\: or \:\:x + 1 = 0$

$\\ \dashrightarrow\:\sf 4x = 0 + 5 \:\:or \:\:x = 0 - 1$

$\\ \dashrightarrow\:\sf 4x = 5 \:\:or \:\:x = -1$

$\\ \dashrightarrow\:\bf \pink{x = \dfrac{5}{4}\:\: or \:\:x = -1}$

$\:$

$\small\therefore\:{\underline{\frak{Zeroes\:of\:given\:polynomial\:\big(\alpha\:and\:\beta\big)\: are \:\bf{\dfrac{5}{4}}\:\frak{and}\:\bf{-1.}}}}$

$\:$

Now,

$\:$

Verifying relationship between zeroes and coefficients of given polynomial ::

$\:$

We know that ::

$\:$

$\quad\odot\:\boxed{\begin{array}{c} \\ \tiny\pink{\bigstar}\:\bf\blue{ Sum\:of\:zeroes\:of\:quadratic\:polynomial = \dfrac{-b}{a}} \\\\ \tiny\pink{\bigstar}\:\bf\green{ Product\:of\:zeroes\:of\:quadratic\:polynomial = \dfrac{c}{a}} \\\\ \end{array}}$

$\:$

So,

$\\ \longrightarrow\:\sf \alpha + \beta = \dfrac{-b}{a},\: \alpha\beta = \dfrac{c}{a}$

$\:$

We have ::

$\:$

• α = 5/4

• β = -1

• a = coefficient of x² = 4

• b = coefficient of x = -1

• c = constant term = -5

$\:$

Putting all known values ::

$\\ \longrightarrow\:\sf \dfrac{5}{4} + \big(-1\big) = \dfrac{-\big(-1\big)}{4},\: \dfrac{5}{4} \:\times\:\big (-1\big) = \dfrac{-5}{4}$

$\\ \longrightarrow\:\sf \dfrac{5}{4} - 1 = \dfrac{1}{4},\: \dfrac{ -5}{4} = \dfrac{-5}{4}$

$\\ \longrightarrow\:\sf \dfrac{5 - 4}{4} = \dfrac{1}{4},\:\dfrac{-5}{4} = \dfrac{-5}{4}$

$\\ \longrightarrow\:\bf \purple{\dfrac{1}{4} = \dfrac{1}{4},\:\dfrac{-5}{4} = \dfrac{-5}{4}}$

$\:$

$\green{\bigstar}$ H E N C E,ㅤV E R I F I E D