Question

Find the value of k if (x - 1) ia a factor of p(x)
a. P(x) = 2x2 + kx + root2
b. P(x) = kx raise to power 2 root2x + 1​

Answer 1

Answer:

Given that ( x - 1 ) is a factor of P(x). Therefore one substituting we must get the value to be zero.

1 ) P(x) = 2x² + kx + √2

⇒ x - 1 = 0

⇒ x = 1

Substituting x = 1 we get,

⇒ 2 ( 1 )² + k ( 1 ) + √2 = 0

⇒ 2 + k + √2 = 0

⇒ k = √2 - 2

2 ) P(x) = kx² + √2 x + 1

Substituting x = 1, we get.

⇒ k ( 1 )² + √2 ( 1 ) + 1 = 0

⇒ k + √2 + 1 = 0

⇒ k = - ( √2 + 1 )

These are the required values.

Hope it helped !!

Answer 2

Answer:

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

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

Step-by-step explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about quadratic whose factor is given.

• We have to find the value of k in both the quadratic.

$\underline \bold{Given : } \\ \implies x - 1 \: is \: a \: factor \: of \: p(x) \\ \implies p(x) = 2 {x}^{2} + kx + \sqrt{2} \\ \implies p(x) = k {x}^{2} + \sqrt{2} x + 1 \\ \\ \underline \bold{To \: Find : } \\ \implies value \: of \: k \: in \: first \:question \\ \implies value \: of \: k \: in \: second \: question$

• According to given question :

$\implies x - 1 = 0 \\ \bold{\implies x = 1} \\ \\\bold{For\:question\:first}\\ \implies p(x) = 2 {x}^{2} + kx + \sqrt{2} \\ \implies p(1) = 2 \times ({1})^{2} + k \times 1 + \sqrt{2} = 0 \\ \implies 2 \times 1 + k + \sqrt{2} = 0 \\ \bold{\implies k = - \sqrt{2} - 2} \\ \\\bold{For\:question\:second} \\ \implies p(x) = k {x}^{2} + \sqrt{2} x + 1 \\ \implies p(1) = k \times {1}^{2} + \sqrt{2} \times 1 + 1 = 0 \\ \implies k + \sqrt{2} + 1 = 0 \\ \bold{\implies k = - \sqrt{2} - 1}$