Question

If αα and ββ are the zeroes of the polynomial f(x)fx = x2+5x+kx2+5⁢x+k such that α−βα-β = −11-11, find the value of kk.​

Answer 1

Answer:-

Given:-

α , β are the roots of x² + 5x + k = 0.

On comparing the given equation with the standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = 1
• b = 5
• c = k.

We know that,

Sum of the roots = - b/a

So, α + β = - 5/1

⟹ α + β = - 5 -- equation (1).

It is given that,

⟹ α - β = - 11 -- equation (2)

Add equations (1) , (2).

⟹ α + β + α - β = - 5 - 11

⟹ 2α = - 16

⟹ α = - 16/2

⟹ α = - 8

Hence, - 8 is one root of given equation.

Substituting x = - 8 in the given equation we get,

⟹ ( - 8)² + 5( - 8) + k = 0

⟹ 64 - 40 + k = 0

⟹ 24 + k = 0

⟹ k = - 24

∴ Value of k is - 24.

Answer 2

$\qquad \qquad \underline {\pmb{\mathbb{ GIVEN \:\:POLYNOMIAL \:\::\:\: } \sf x^2 + 5x + k \:}}$

As, We know that ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Sum \:of \:zeroes \:\:\red {\:( \:\alpha \: + \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{\:-(Cofficient \:of \:x\:)\:\:}{Cofficient \:of \:x^2 \:}$

⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \bigg( \alpha +\:\beta \: \bigg) \:=\:\dfrac{-(\:Cofficient \:of \:x\:)}{Cofficient \:of \:x^2 \:}\\\\\\\qquad \dashrightarrow \sf \Big\{ \alpha \:+ \beta\: \Big\} \:=\:\dfrac{-(\:5\:)}{1 \:}\\\\\\\qquad \dashrightarrow \sf \Big\{ \alpha \:+ \beta\:\: \Big\} \:=\:\dfrac{–5}{1} \\\\\\\qquad \dashrightarrow \sf \alpha \:+ \beta\: \: \:=\:– \:5\\\\\\ \dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \:+ \beta\: \: \:=\:– \:5\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

Given that ,

$\qquad \dashrightarrow \sf \alpha \:- \beta\: \: \:=\:– \:11 \:\\\\\\ \qquad \dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \:- \beta\: \: \:=\:– \:11\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀¤ Finding Value of : α

$\qquad :\implies \sf \alpha + \beta \: \:+\: \alpha - \beta \:\:\\\\\\\qquad :\implies \sf \Big\{ \alpha + \beta \:\Big\} \:+\:\Big\{ \alpha - \beta \:\Big\}\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

• α + β = – 5 &
• α – β = – 11

$\qquad :\implies \sf \Big\{ \alpha + \beta \:\Big\} \:+\:\Big\{ \alpha - \beta \:\Big\}\: =\:\Big\{ – 5 \:\Big\} \:+\:\Big\{ – 11 \:\Big\}\: \\\\\\ :\implies \sf 2 \alpha\: =\:– \:16 \:\: \\\\\\ :\implies \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \: \: \:=\:– \:8\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀¤ Finding Value of : k

As , We know that ,

• α is one of the root of the polynomial : x² + 5x + k

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Value \:of \: \alpha \:(\:or \:x\:) \::}}$

$\qquad :\implies \sf x^2 + 5x + k \:\\\\\\ \qquad :\implies \sf (-8)^2 + 5(-8) + k \:=\:0\:\\\\\\ \qquad :\implies \sf 64 + (-40) + k \:=\:0\:\\\\\\ \qquad :\implies \sf 64 - 40 + k \:=\:0\:\\\\\\ \qquad :\implies \sf 24 + k \:=\:0\:\\\\\\ \qquad :\implies \sf k \:=\:-24\:\\\\\\ :\implies \underline{\boxed{\purple{\pmb{\frak{ \:k \: \: \:=\:– \:24\: }}}}}\:\:\bigstar$

$\qquad \therefore \underline {\:\sf Hence, \: The \:Value \:of \: k \:\pmb{\bf –\:24 \:}\:.}$