Verified answer

√(2x+14) > x+3

[√(2x + 14)]^2 > (x + 3)^2

2x + 14 > x^2 + 6x + 9

x^2 + 4x - 5 < 0

(x + 5)(x - 1) < 0

-5 < x < 1

But the domain is:

(2x + 14) ≥ 0

2x ≥ -14

x ≥ -7

So combining the two case we get:

-7 ≤ x < 1

find where they intercept eachother then check points on either sides of the intercepts to find where the statement √(2x+14) > x+3 holds true.

to find where they intersect make them equal to eachother.

√(2x+14) = x+3

2x + 14 = (x+3)^2

2x + 14 = x^2 + 6x + 9

0 = x^2 + 4x - 5

0 = (x + 5)(x - 1)

x = -5, x = 1

check for extraneous solutions by substituting x= -5 and x = 1 into the equation √(2x+14) = x+3

√(2(-5)+14) = -5+3

2 = -2 this is not true therefore x = -5 is not a solution

√(2(1)+14) = 1+3

4 = 4 this is true therefore x = 1 is a solution.

so now we know they intercept when x = 1

now just test a point on either side of x = 1 to find where √(2x+14) > x+3 is true

we have infact already checked x = -5 when we were checking for extraneous solution. so for values to the left of x = 1 the statement √(2x+14) > x+3 holds true. if you test a point to the right of x = 1 you will find that the statement is not true.

so we know x < 1

but there is also a lower limit to x because we cant take square root of a negative number.

√(2x+14) = 0

2x + 14 = 0

2x = -14

x = -7

so the final answer is

-7 ≤ x < 1

Square both sides,

2x + 14 > (x + 3)²

2x + 14 > x² + 6x + 9

x² + 4x - 5 < 0

Now factor,

(x - 1)(x + 5) < 0

-5 < x < 1

However, the squaring introduced a spurious solution to √(2x+14) = x+3

When -7 < x ≤ -5 the left side is positive but the right side is negative in the original inequality. When x = -7 the left side is 0 and the right is negative.

When x < -7 you will be taking the square root of a negative number. Therefore the solution is,

-7 ≤ x < 1

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