Question

Find the Zeroes of the following quadratic polynomial and verify the relationship between
the zeroes and the coefficients

x2 - 2x - 8​

Answer 1

$\large\sf \underline{ \color{red}{Answer : -}}$

$\rm \sf{Zeroes \: of \: the \: quadratic \: polynomial \: {x}^{2} - 2x - 8 \: is \: - 4 \: and \: 2}$

${ \boxed {\sf{ \color{purple}{solution : }}}}$

$\rm{given \: equation \: {x}^{2} - 2x - 8}$

• $\leadsto$calculate the Zeroes of the polynomial
• $\leadsto$ Apply the method of splitting the middle term

$\sf{ : \implies \: {x}^{2} - (4x - 2x) - 8 = 0}$

$\sf{ : \implies \: {x}^{2} - 4x + 2x - 8 = 0}$

$\sf{ : \implies \: x(x - 4) + 2(x - 4) = 0}$

$\sf{ : \implies \: ( x - 4)(x + 4) = 0}$

$\sf{ : \implies \: x = 4 \:, x = - 2}$

• Therefore , two Zeroes of this polynomial are 4 , - 2
• Now ,

• Verífy the Zeroes as the sum of zeroes and product of Zeroes.

$\sf \leadsto \: { {x}^{2} - 2x - 8} \: \: \: \: \: \: \: \: .. \: \tiny{eq.1}$

• Comparing equation 1 with general form of quadratic polynomial , then

$\sf{(a = 1),(b = - 2),(c = - 8)}$

• sum of zeroes = LHS

$\mapsto \:{\boxed{ \sf{ = \: sum \: of \: zeroes \: = \alpha + \beta} }}$

$\circ \: \sf{substitute \: the \: zeroes \: (\alpha = 4 )\: and \: (\beta = - 2)}$

$\longrightarrow \sf {LHS = 4 + ( - 2 ) = 2 }$

$\longrightarrow \sf {RHS = \frac{ - b}{c} }$

• substitute the values a = 1 and b = -2

$\leadsto \: \sf{ \frac{ - b}{a} = \frac{ - ( - 2)}{1}}$

$\leadsto \: \sf{ \frac{ - b}{a} = 2}$

$\sf{ \therefore \: LHS = RHS}$

• product of zeroes = LHS

$\mapsto \:{\boxed{ \sf{ = product \: of \: zeroes \: = \alpha \times \beta }}}$

$\circ \: \sf{substitute \: the \: zeroes \: (\alpha = 4 )\: and \: (\beta = - 2)}$

$\implies \: \: \sf{( \alpha = 4) \: and \: ( \beta = - 2)}$

$\leadsto \sf{4 \times ( - 2)}$

$\leadsto { \boxed{\sf{ - 8}}}$

$\longrightarrow \sf {RHS = \frac{c}{a} }$

• substitute the values a = 1 and c = - 8

$\leadsto \: \sf{ \frac{c}{a} = \frac{ - ( - 8)}{1}}$

$\leadsto \: \sf{ \frac{c}{a} = - 8}$

$\sf{ \therefore \: LHS = RHS}$

Hence

the sum of Zeroes and product of Zeroes are verified .

$\circ \: { \boxed{ \color{red}\sf{Zeroes \: of \: quadratic \: polynomial \: \: {{x}^{2} - 2x - 8 \: is \: - 4 , \: 2}}}}$

Answer 2

♡ AnSweR ♡

✯..x2 - 2x - 8..✯

✯..f(x)=x 2 −2x−8

⇒f(x)=x 2−4x+2x−8

⇒f(x)=x(x−4)+2(x−4)]

⇒f(x)=(x−4)(x+2)

✯..Zeros of f(x) are given by f(x) = 0

⇒x 2 −2x−8=0

⇒(x−4)(x+2)=0

⇒x=4 or x=−2

So, α=4 and β=−2

∴ sum of zeros =α+β=4−2=2

Also, sum of zeros = Coefficient of x

Coefficient of x²

= -(-2) =2

1

So, sum of zeros =α+β= −Coefficient of x2

Coefficient of x

Now, product of zeros =αβ=(4)(−2)=−8

Also, product of zeros = Constant terms

Coefficient of x²

= 1−8 =−8

∴ Product of zeros = Coefficient of x2

Constant term

=αβ