What is the answer to solvefor a,f(a+2)=sqrt(2+a)+8 ?

The answer to solvefor a,f(a+2)=sqrt(2+a)+8 is a=(-4f^2+16f+1+sqrt(32f+1))/(2f^2) {f>0}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor a,f(a+2)=sqrt(2+a)+8

solve for

a, f (a + 2) = 2 + a + 8

a = \frac{- 4 f ^{2} + 16 f + 1 + 32 f + 1}{2 f ^{2}} {f > 0}

Solution steps

f (a + 2) = 2 + a + 8

Expand

f (a + 2) : a f + 2 f

a f + 2 f = 2 + a + 8

Remove square roots

a^{2} f^{2} + 4 a f^{2} - 16 a f + 4 f^{2} - 32 f + 64 = 2 + a

Solve

a^{2} f^{2} + 4 a f^{2} - 16 a f + 4 f^{2} - 32 f + 64 = 2 + a : a = \frac{- 4 f ^{2} + 16 f + 1 + 32 f + 1}{2 f ^{2}}, a = \frac{- 4 f ^{2} + 16 f + 1 - 32 f + 1}{2 f ^{2}}

a = \frac{- 4 f ^{2} + 16 f + 1 + 32 f + 1}{2 f ^{2}}, a = \frac{- 4 f ^{2} + 16 f + 1 - 32 f + 1}{2 f ^{2}}

Verify Solutions:

a = \frac{- 4 f ^{2} + 16 f + 1 + 32 f + 1}{2 f ^{2}} {f > 0}, a = \frac{- 4 f ^{2} + 16 f + 1 - 32 f + 1}{2 f ^{2}}

False

The solution is

a = \frac{- 4 f ^{2} + 16 f + 1 + 32 f + 1}{2 f ^{2}} {f > 0}

x^{\frac{1}{3}} + 10 = 12

19 - 3 x = 5

3 - 15 - 2 x - x^{2} = 0

x^{2} + 48 x = 10

What is the answer to solvefor a,f(a+2)=sqrt(2+a)+8 ?
The answer to solvefor a,f(a+2)=sqrt(2+a)+8 is a=(-4f^2+16f+1+sqrt(32f+1))/(2f^2) {f>0}