en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,f(x-3)=x^2-1

Solution

solve for

x, f (x - 3) = x^{2} - 1

Solution

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

Solution steps

f (x - 3) = x^{2} - 1

Expand

f (x - 3) : x f - 3 f

x f - 3 f = x^{2} - 1

Switch sides

x^{2} - 1 = x f - 3 f

Move

3 f

to the left side

x^{2} - 1 + 3 f = x f

Move

x f

to the left side

x^{2} - 1 + 3 f - x f = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - f x - 1 + 3 f = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - f ) \pm ( - f ) ^{2} - 4 \cdot 1 \cdot ( - 1 + 3 f )}{2 \cdot 1}

Simplify

(- f)^{2} - 4 \cdot 1 \cdot (- 1 + 3 f) : f^{2} - 4 (- 1 + 3 f)

x_{1, 2} = \frac{- ( - f ) \pm f ^{2} - 4 ( - 1 + 3 f )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - f ) + f ^{2} - 4 ( - 1 + 3 f )}{2 \cdot 1}, x_{2} = \frac{- ( - f ) - f ^{2} - 4 ( - 1 + 3 f )}{2 \cdot 1}

x = \frac{- ( - f ) + f ^{2} - 4 ( - 1 + 3 f )}{2 \cdot 1} : \frac{f + f ^{2} - 4 ( - 1 + 3 f )}{2}

x = \frac{- ( - f ) - f ^{2} - 4 ( - 1 + 3 f )}{2 \cdot 1} : \frac{f - f ^{2} - 4 ( - 1 + 3 f )}{2}

The solutions to the quadratic equation are:

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

Popular Examples

solvefor x,f(x-3)=x^2-2 solve for

x, f (x - 3) = x^{2} - 2

x^{2} - 529 = 0

2 x + 4 = (x - 2)^{2}

x^{2} - 11 x + 28 = 0, x

0=-16t^2+16t+64

0 = - 16 t^{2} + 16 t + 64