What is sqrt(x^2-4x+13)+sqrt(x^2-6x+13)>= 5 ?

The solution to sqrt(x^2-4x+13)+sqrt(x^2-6x+13)>= 5 is True for all x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(x^2-4x+13)+sqrt(x^2-6x+13)>= 5

x^{2} - 4 x + 13 + x^{2} - 6 x + 13 \geq 5

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

x^{2} - 4 x + 13 + x^{2} - 6 x + 13 \geq 5

Subtract

x^{2} - 6 x + 13

from both sides

x^{2} - 4 x + 13 + x^{2} - 6 x + 13 - x^{2} - 6 x + 13 \geq 5 - x^{2} - 6 x + 13

Simplify

x^{2} - 4 x + 13 \geq 5 - x^{2} - 6 x + 13

f (x) \geq g (x)

then

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

For

5 - x^{2} - 6 x + 13 < 0 : x < - 21 + 3 or x > 21 + 3

For

5 - x^{2} - 6 x + 13 \geq 0 : - 21 + 3 \leq x \leq 21 + 3

Combine the intervals

- 21 + 3 \leq x \leq 21 + 3 or x < - 21 + 3 or x > 21 + 3

Merge Overlapping Intervals

True for all x

Verify Solutions:

The solution is

True for all x

(4 - x) (x - 3)^{\frac{1}{2}} < 0

\frac{( x - 1 )}{( x + 1 )} > 0

x - 1 > 3 - x

x^{2} - 5 x + 4 \leq 2 x - 2

2 x - 3 - x - 2 > 2

What is sqrt(x^2-4x+13)+sqrt(x^2-6x+13)>= 5 ?
The solution to sqrt(x^2-4x+13)+sqrt(x^2-6x+13)>= 5 is True for all x