What is sqrt(x^2-3x+2)>2x-5 ?

The solution to sqrt(x^2-3x+2)>2x-5 is x<= 1\lor 2<= x<(sqrt(13)+17)/6

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sqrt(x^2-3x+2)>2x-5

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

x \leq 1 or 2 \leq x < \frac{13 + 17}{6}

Interval Notation

(- \infty, 1] \cup [2, \frac{13 + 17}{6})

Decimal Notation

x \leq 1 or 2 \leq x < 3.43425 \dots

Solution steps

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

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

2 x - 5 < 0 : x < \frac{5}{2}

For

2 x - 5 \geq 0 : \frac{5}{2} \leq x < \frac{13 + 17}{6}

Find singularity points

Find non-negative values for radicals:

x \leq 1 or x \geq 2

Combine the intervals

(\frac{5}{2} \leq x < \frac{13 + 17}{6} or x < \frac{5}{2}) and (x \leq 1 or x \geq 2)

Merge Overlapping Intervals

x \leq 1 or 2 \leq x < \frac{13 + 17}{6}

Verify Solutions:

x \leq 1

True

, 2 \leq x < \frac{13 + 17}{6}

True

The solution is

x \leq 1 or 2 \leq x < \frac{13 + 17}{6}

x - 4 < 2

(x^{2} - 1) \cdot x^{2} - x - 2 \geq 0

\frac{( 1 - 21 - 4 x - x ^{2} )}{( x + 1 )} \geq 0

x - 1 - ∣ x ∣ \leq 0

3 - x \leq 1 - x

What is sqrt(x^2-3x+2)>2x-5 ?
The solution to sqrt(x^2-3x+2)>2x-5 is x<= 1\lor 2<= x<(sqrt(13)+17)/6