The solution to x^2-4x+3>x is x (sqrt(13)+5)/2

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x^2-4x+3>x

Solution

x^{2} - 4 x + 3 > x

x < \frac{- 13 + 5}{2} or x > \frac{13 + 5}{2}

Interval Notation

(- \infty, \frac{- 13 + 5}{2}) \cup (\frac{13 + 5}{2}, \infty)

Decimal Notation

x < 0.69722 \dots or x > 4.30277 \dots

Solution steps

x^{2} - 4 x + 3 > x

Rewrite in standard form

x^{2} - 5 x + 3 > 0

Complete the square

x^{2} - 5 x + 3 : (x - \frac{5}{2})^{2} - \frac{13}{4}

(x - \frac{5}{2})^{2} - \frac{13}{4} > 0

Move

\frac{13}{4}

to the right side

(x - \frac{5}{2})^{2} > \frac{13}{4}

For

u^{n} > a

, if

n

is even then

x - \frac{5}{2} < - \frac{13}{4} or x - \frac{5}{2} > \frac{13}{4}

x - \frac{5}{2} < - \frac{13}{4} : x < \frac{- 13 + 5}{2}

x - \frac{5}{2} > \frac{13}{4} : x > \frac{13 + 5}{2}

Combine the intervals

x < \frac{- 13 + 5}{2} or x > \frac{13 + 5}{2}

\frac{9 ^{- 3}}{9 ^{6}}

x^{2} + 8 x - 60 > 7 x - 4

- 5 c = 35

d - 4 > 3 or d - 4 \leq 1

s^{6} - t^{3}

What is x^2-4x+3>x ?
The solution to x^2-4x+3>x is x<(-sqrt(13)+5)/2 \lor x>(sqrt(13)+5)/2