What is sqrt(x^2-x-12)<= sqrt(x^2-6x+5) ?

The solution to sqrt(x^2-x-12)<= sqrt(x^2-6x+5) is x<=-3

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sqrt(x^2-x-12)<= sqrt(x^2-6x+5)

x^{2} - x - 12 \leq x^{2} - 6 x + 5

x \leq - 3

Interval Notation

(- \infty, - 3]

Solution steps

x^{2} - x - 12 \leq x^{2} - 6 x + 5

Square both sides

(x^{2} - x - 12)^{2} \leq (x^{2} - 6 x + 5)^{2}

Simplify

x^{2} - x - 12 \leq x^{2} - 6 x + 5

x^{2} - x - 12 \leq x^{2} - 6 x + 5 : x \leq \frac{17}{5}

x \leq \frac{17}{5}

Find singularity points

Find non-negative values for radicals:

x \leq - 3 or x \geq 5

Combine the intervals

x \leq \frac{17}{5} and (x \leq - 3 or x \geq 5)

Merge Overlapping Intervals

x \leq - 3

6 x = - 4

x^{2} + 6 x y - 16 y^{2}

x^{2} - x = 56

- 4 x = - 2

3 x^{2} + 2 x + 4 = 0

What is sqrt(x^2-x-12)<= sqrt(x^2-6x+5) ?
The solution to sqrt(x^2-x-12)<= sqrt(x^2-6x+5) is x<=-3