What is sqrt(x^2-6x+8)<= sqrt(x+1) ?

The solution to sqrt(x^2-6x+8)<= sqrt(x+1) is (-sqrt(21)+7)/2 <= x<= 2\lor 4<= x<= (sqrt(21)+7)/2

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sqrt(x^2-6x+8)<= sqrt(x+1)

x^{2} - 6 x + 8 \leq x + 1

\frac{- 21 + 7}{2} \leq x \leq 2 or 4 \leq x \leq \frac{21 + 7}{2}

Interval Notation

[\frac{- 21 + 7}{2}, 2] \cup [4, \frac{21 + 7}{2}]

Decimal Notation

1.20871 \dots \leq x \leq 2 or 4 \leq x \leq 5.79128 \dots

Solution steps

x^{2} - 6 x + 8 \leq x + 1

Square both sides

(x^{2} - 6 x + 8)^{2} \leq (x + 1)^{2}

Simplify

x^{2} - 6 x + 8 \leq x + 1

x^{2} - 6 x + 8 \leq x + 1 : \frac{- 21 + 7}{2} \leq x \leq \frac{21 + 7}{2}

\frac{- 21 + 7}{2} \leq x \leq \frac{21 + 7}{2}

Find singularity points

Find non-negative values for radicals:

- 1 \leq x \leq 2 or x \geq 4

Combine the intervals

\frac{- 21 + 7}{2} \leq x \leq \frac{21 + 7}{2} and (- 1 \leq x \leq 2 or x \geq 4)

Merge Overlapping Intervals

\frac{- 21 + 7}{2} \leq x \leq 2 or 4 \leq x \leq \frac{21 + 7}{2}

\frac{( x - 1 )}{( x - 2 )} < 0

\frac{( x ^{2} - 13 x + 40 )}{( 19 x - x ^{2} - 78 )} \leq 0

x - 2 \geq - 1

x - 1 - x < 0

x^{2} + x - 12 < x

What is sqrt(x^2-6x+8)<= sqrt(x+1) ?
The solution to sqrt(x^2-6x+8)<= sqrt(x+1) is (-sqrt(21)+7)/2 <= x<= 2\lor 4<= x<= (sqrt(21)+7)/2