What is 5+sqrt(c-3)<= 6 ?

The solution to 5+sqrt(c-3)<= 6 is 3<= c<= 4

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5+sqrt(c-3)<= 6

5 + c - 3 \leq 6

3 \leq c \leq 4

Interval Notation

[3, 4]

Solution steps

5 + c - 3 \leq 6

Move

5

to the right side

c - 3 \leq 1

Square both sides

(c - 3)^{2} \leq 1^{2}

Simplify

c - 3 \leq 1

Move

3

to the right side

c \leq 4

Find singularity points

Find non-negative values for radicals:

c \geq 3

Combine the intervals

c \leq 4 and c \geq 3

Merge Overlapping Intervals

3 \leq c \leq 4

144 a^{2} b^{4} c^{6}

8 p^{2} + 7 p

1 = x^{2} + 2 x - 1

x^{2} - 20 x + 75