Question

Question:-
Find the value of k, if x - 1 is a factor of p(x) of the question below:-
i) p(x) = kx² - $\sqrt{2x}$ + 1

Note:-
▪️Grade - 9
▪️Chapter - Polynomials
✅Quality answer needed
✅Step - by - step explanation required ( must )
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Answer 1

Answer:

Finding g(x)

If ,

x-1 = 0

x = 0 +1                        (transposing 1)

x = 1

p(x) = kx² -  + 1

Substituting the value of x as (-1)

p(1) = k(-1)² - $\sqrt{2}$ x (1)+ 1

p(1) = k -  $\sqrt{2}$ +1

k= $\sqrt{2}$ -1          (transposing k)

We can leave the answer as  $\sqrt{2}$ -1 without considering the values without substitution.. 

Verification :

($\sqrt{2}$- 1) x² - ($\sqrt{2x}$+1) = 0

($\sqrt{2}$- 1) (1) -  ($\sqrt{2x}$+1)  = 0

($\sqrt{2}$- 1) - ($\sqrt{2x}$+1)  = 0

0 = 0

LHS = RHS

Answer 2

Given that , x - 1 is a factor of p(x) of the question below ; (i) p(x) = kx² - $\sqrt{2x}$ + 1 .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's say the factor x - 1 be g(x) .

$\qquad \therefore \sf \:g(x) \: =\: x \:-\: 1 \:=\: 0 \:$

$\qquad \dashrightarrow \sf \:g(x) \: =\: x \:-\: 1 \:=\: 0 \:$

$\qquad \dashrightarrow \sf \:g(x) \: =\: x \: \:=\: 1 \:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{\:g(x) \: =\:x \:=\: 1 \:}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀¤ Finding Value of k :

$\qquad \dashrightarrow \sf kx^2 \:-\:\sqrt{2}x \: + 1 \:$

⠀⠀⠀⠀⠀Given that ,

• x - 1 is a factor of p(x) of the question below ; (i) p(x) = kx² - $\sqrt{2x}$ + 1 .

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:g(x) \: =\:x \:=\: 1 \: }\bigg\rgroup$

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

$\qquad \dashrightarrow \sf kx^2 \:-\:\sqrt{2}x \: + 1 \:= \:0$

$\qquad \dashrightarrow \sf k(1)^2 \:-\:\sqrt{2}(1) \: + 1 \:=\:0$

$\qquad \dashrightarrow \sf k(1) \:-\:\sqrt{2} \: + 1 \:=\:0$

$\qquad \dashrightarrow \sf k \:-\:\sqrt{2} \: + 1 \:=\:0$

$\qquad \dashrightarrow \sf k \: \:=\:\sqrt{2} - 1 \:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{\: k \: \:=\:\sqrt{2} - 1 \:\:}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline {\underline {\mathbb{\:\:\bigstar \:\:VERIFICATION \:\::\:}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀G I V E N P O L Y N O M I A L : p(x) = kx² - $\sqrt{2}$x + 1

$\qquad \dashrightarrow \sf kx^2 \:-\:\sqrt{2}x \: + 1 \:= \:0$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:g(x) \: =\:x \:=\: 1 \: }\bigg\rgroup$

$\qquad \dag\:\:\bigg\lgroup \sf{ \:k \: \:=\:\sqrt{2} - 1 \: \: }\bigg\rgroup$

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

$\qquad \dashrightarrow \sf kx^2 \:-\:\sqrt{2}x \: + 1 \:= \:0$

$\qquad \dashrightarrow \sf \Big\{ \sqrt{2} - 1 \Big\} \:(1)^2 \:-\:\sqrt{2}(1) \: + 1 \:= \:0$

$\qquad \dashrightarrow \sf \Big\{ \sqrt{2} - 1 \Big\} \:(1) \:-\:\sqrt{2} \: + 1 \:= \:0$

$\qquad \dashrightarrow \sf \sqrt{2} - 1 \:-\:\sqrt{2} \: + 1 \:= \:0$

$\qquad \dashrightarrow \sf 0 \:= \:0$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{\: 0 \: \:=\:0 \:\:}}}}}$

⠀⠀⠀⠀⠀$\therefore {\underline {\pmb{\bf Hence, \:Verified \:}}}$