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Popular >

solvefor x,f(x-2)=sqrt(x-3)

Solution

solve for

x, f (x - 2) = x - 3

Solution

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

Solution steps

f (x - 2) = x - 3

Expand

f (x - 2) : x f - 2 f

x f - 2 f = x - 3

Square both sides:

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

Solve

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3 : x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}}

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}}

Verify Solutions:

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

The solutions are

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

Popular Examples

sqrt(x^2+2x)=sqrt(27-4x)

x^{2} + 2 x = 27 - 4 x

5x+4=(x+4)^{1/2}+2

5 x + 4 = (x + 4)^{\frac{1}{2}} + 2

sqrt(2x-5)-sqrt(3x+4)=1

2 x - 5 - 3 x + 4 = 1

x + 5 x = 12

18sqrt(8)-8sqrt(18)=sqrt(n)

188 - 818 = n

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