Question

Find the value of k, if x-1 is a factor of p(x) in each of the following cases.
$(i). \: p(x) = k {x}^{2} - \sqrt{2} x + 1$
$(ii). \: p(x) = k {x}^{2} - 3x + k$
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Answer 1

Answer:

p(x) = kx² - 3x + k

If (x - 1) is a factor of p(x)

Therefore

x - 1 = 0

x = 1

Substitute the value of x in Equation

⇒ k(1)² - 3 × 1 + k = 0

⇒ 2k - 3 = 0

⇒ 2k = 3

⇒ k =3/2

p(x) = kx² - √2x + 1

Now x - 1 is a factor

x - 1 = 0

x = 1

Now put x = 1 in p(x)

p(1) = 0

p(1) = k(1)² - √2(1) + 1 = 0

k - √2 + 1 = 0

k = √2 - 1

So required value of k = √2 - 1

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Step-by-step explanation:

Answer 2

ANSWER:

1) P(x)=kx²-√2x+1

As x-1 is a factor of p(x)

.: by factor theorem

x-1=0

x=0

=> p(1)=0

=> k(1)²-√2(1)+1=0

=> k-√2+1= 0

=> k = √2-1

2) P(x)=kx²-3x+k

As x-1 is a factor of p(x)

.: by factor theorem

x-1=0

x= 1

=> p(1)=0

=> k(1)²-3(1)+k=0

=> k-3+k = 0

=> 2k = 3

=> k= 3/2

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