Question

P(x)=2x square +kx+5 if p(-1)=0 let K

Answer 1

Question:

If p(x) = 2x^2 + kx + 5 and p(-1) = 0,

then find the value of "k" .

Answer:

k = 7

Solution:

Given,

p(x) = 2x^2 + kx + 5 --------(1)

p(-1) = 0 ----------(2)

Also,

If p(x) is given, then to find the value of p(a) , we just need to put x = a , everywhere in the expression of p(x) .

Thus,

If p(x) = 2x^2 + kx + 5

Then,

=> p(-1) = 2(-1)^2 + k•(-1) + 5

=> p(-1) = 2•1 - k + 5

=> p(-1) = 2 - k + 5

=> p(-1) = 7 - k

=> 0 = 7 - k {using eq-(2) ,p(-1) = 0}

=> k = 7

Hence,

The required value of "k" is 7 .

Answer 2

Question :

$\small p(x) = 2 {x}^{2} + kx + 5 \: \: \: if \: \: p( - 1) = 0 \\ \\ then \: find \: the \: value \: of \: k \: .$

Answer:

$\implies \: \green{\boxed{ \red{ \boxed{k = 7}}}}$

Explanation:

if any equation of given p(x) ,to find any values of p(n) put x = n in the given equation .

Solution :-

we have,

$p(x) = 2 {x}^{2} + kx + 5$

Given that,

$p \: ( - 1) = 0 \\ \\ \therefore \: put \: \: x \: = - 1 \\ \\ \small \:p ( - 1) = 0 = 2 \times {( - 1)}^{2} + k \times (- 1) + 5 \\ \\ \implies \: 0 = 2 - k + 5 \\ \\ \implies \: 0 = 7 - k \\ \\ \implies \: \: - 7 = - k \\ \\ \implies \: \red{ \boxed{k = 7}}$