Question

The value of b if f (x) =x²+4√x+b and f(16)=275​

Answer 1

Answer:

put value of

$x = 16$

in

${x}^{2} + 4 \sqrt{x} + b \\ 256 + 4 \sqrt{16} + b \\ 256 + 16 + b = 275$

Answer 2

Answer:

The value of b is 3.

Step-by-step explanation:

Given,

$f(x) = {x}^{2} + 4 \sqrt{x} + b$

Here also given value of f(16).

We want to find value of b.

We are putting x=16 and get,

$f(16) = {16}^{2} + 4 \sqrt{16} + b \\ = 256 + 4 \times 4 + b \\ = 256 + 16 + b \\ = 272 + b$

According to question,

$272 + b = 275 \\ b = 275 - 272 \\ b = 3$

Required value of b is 3.

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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2) https://brainly.in/question/1169549