Question

Find the Zeroes of the following quadratic
polynomial x² 2X-8 and verify the relationship
between the teroes and the to woefficients​

Answer 1

Solution :

Here, given a quadratic polynomial x² - 2x - 8

We've to find it's zeroes and verify the relationship between it's zeroes and coefficients.

We'll find it's zeroes by splitting the middle term.

Now

➥ x² - 2x - 8 = 0

➥ x² - 4x + 2x - 8 = 0

➥ x (x - 4) + 2 (x - 4) = 0

➥ (x + 2) (x - 4)

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➝ x + 2 = 0

➝ x = -2

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➝ x - 4 = 0

➝ x = 4

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[.°. x = -2, 4 ]

Sum of zeroes :

➥ (-2) + 4

➥ 4 - 2

➥ 2

Product of zeroes :

➥(-2) × 4

➥ -8

[ α + β = 2 ; αβ = -8 ]

Now, we'll verify the relationship between zeroes and it's coefficients.

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Let "a" be the coefficient of x², "b" be the coefficient of x and "c" be the constant term

Substituting the values i.e. a = 1 , b = -2 and c = -8.

• α + β = -b/a = -(-2)/1 => 2
• αβ = c/a = -8/1 => -8

Hence, verified !

Answer 2

${\huge{\rm{\underbrace{Correct\; Question:-}}}}$

Q) Find the zeroes of the Quadratic Polynomial :

→ x² - 2x - 8

and verify the relationship between the zeroes and the coefficients .

${\huge{\rm{\underbrace{Answer\checkmark}}}}$

Analysis :

We are given with a Quadratic Polynomial so , we must know that a Quadratic Polynomial has 2 roots or zeroes . The roots/zeroes of the Polynomial can be found easily either by doing Middle Term Splitting or by using Quadratic Formula .

After finding the zeroes , we will establish the relationship of Sum of Zeroes and Product of Zeroes with the coefficients of the Polynomial .

${\textit{\textbf{Finding\;the\;Zeroes:}}}$

We would use Middle Term Splitting to find the Zeroes .

$\rm \rightarrow \: {x}^{2} - 2x - 8 = 0$

$\rm \rightarrow \: {x}^{2} -4x + 2x - 8 = 0$

$\rightarrow \rm \: x(x - 4) + 2(x - 4) = 0$

$\rightarrow \rm \: (x - 4)(x + 2) = 0$

So ,

Either

$\mapsto \rm \: x - 4 = 0 \\ \mapsto \: \: \: \underline{\boxed{ \purple{ \rm{x = 4}}}}$

or

$\mapsto \rm \: x + 2 = 0 \\ \mapsto \: \: \: \: \: \: \underline{ \boxed{ \rm{ \pink{x =- 2}}}}$

So ,

Zeroes of the Polynomial :

$\green{ \star} \: \: \: \: \: \: \: \boxed{ 4 \: \rm \: and \: \: -2 \: }$

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$\boxed{ \sf{ {x}^{2} - 2x - 8}}$

compare this equation with the standard Quadratic Equation .

$\bull \rm \: \: \: {x}^{2} - 2x - 8 \: \rang \lang \: a{x}^{2} + bx + c$

So ,

• a = 1
• b = -2
• c = -8

We know ,

$\mapsto \rm \: sum \: of \: zeroes = \frac{ - b}{a}$

Here ,

LHS = 4 + (-2) = 4 - 2 = 2 .

RHS = $\dfrac{-(-2)}{1} = \bf2$

LHS = RHS (Hence Verified)

Now ,

$\mapsto \rm \: product \: of \: zeroes \: = \frac{c}{a}$

Here ,

LHS = 4 × -2 = -8

RHS = $\dfrac{-8}{1}=\bf{-8}$

LHS = RHS (Hence Verified)

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