Solve Inequalities involving root functions

We are often required to solve inequalities involving the square roots of unknowns, and there are various techniques for solving these. There are algebraic methods, and there are graphical methods, and we will demonstrate some of these here.

The most important thing to remember, when solving these types of inequalities, is to recognise that the quantity under the square root sign must always remain greater than or equal to zero. This should always be your first observation.

Any algebraic solution you arrive at must also satisfy the first observation. So for example, an algebraic simplification might deliver the solution given by $22<x<10. However, suppose you find that to keep the expression under the square root sign positive or zero, you require that $x\ge4$x≥4. Your conclusion therefore becomes the intersection of these inequalities. That is, the correct solution is given by $4\le x<10$4≤x<10.     

Study the following six examples carefully.

Example 1: 

Solve $\sqrt{x}>5$√x>5

This is not difficult. Clearly, by observation, we must have $x>25$x>25, since $\sqrt{25}=5$√25=5, so anything greater than $25$25 will be a solution to $\sqrt{x}>5$√x>5.

However, if we proceed algebraically using the suggested strategy, we need to first observe that the quantity under the square root must be greater than or equal to $0$0. 

Thus we start with recognising that $x\ge0$x≥0. No matter what our algebra offers us as a solution, it must be seen in the light of this overriding restriction on $x$x. Think of the interval $x\ge0$x≥0  as the allowable space for which any solution to the question must sit in. 

So we  commence to solve the question using algebra. This involves squaring both sides of the inequality:

$\sqrt{x}$√x	$>$>	5
x	$>$>	25

We then look at our derived solution and check that it lies within the  interval $x\ge0$x≥0. Of course, it does, so we correctly write down our solution as $\left\{x:x>25,x\in\Re\right\}${x:x>25,x∈ℜ}.

Example 2:

Solve $\sqrt{x+7}\ge12$√x+7≥12

Our first observation is that the quantity $x+7$x+7 must be greater than or equal to $0$0. This means that $x\ge-7$x≥−7. Any solution from algebra cannot be outside this interval.

Again, we solve our equation, using the strategy of squaring:

$\sqrt{x+7}$√x+7	$>=$>=	$12$12
$x+7$x+7	$>=$>=	$144$144
$x$x	$>=$>=	$137$137

Our solution is well inside the interval $x\ge-7$x≥−7 so we accept the solution as $\left\{x:x\ge137,x\in\Re\right\}${x:x≥137,x∈ℜ}.

Example 3:

Solve $\sqrt{3x-6}<6$√3x−6<6

We observe first that $3x-6>=0$3x−6>=0, and this means that $3x>=6$3x>=6, or that $x>=2$x>=2.

Now solving:

$\sqrt{3x-6}$√3x−6	$<$<	$6$6
$3x-6$3x−6	$<$<	$36$36
$3x$3x	$<$<	$42$42
$x$x	$<$<	$14$14

Since $x>=2$x>=2, then the solution is not just $x<14$x<14  because that includes numbers less than $2$2 as well. Rather, the correct solution becomes $\left\{x:2\le x<14,x\in\Re\right\}${x:2≤x<14,x∈ℜ}. 

Example 4:

Solve $11-3\sqrt{x}\le-1$11−3√x≤−1

Here, check that we must ensure that $x\ge0$x≥0.

We proceed to solve, taking care that, if we multiply or divide both sides of an inequality by a negative number, the inequality sign reverses direction. Thus:

$11-3\sqrt{x}$11−3√x	$<=$<=	$-1$−1
$$	$<=$<=	$-12$−12
$$	$>=$>=	$4$4
$x$x	$>=$>=	$16$16

Our solution poses no issue at all, because every value of $x$x greater than or equal to $16$16  also satisfies $x>=0$x>=0. So our formal solution is given by $\left\{x:x\ge16,x\in\Re\right\}${x:x≥16,x∈ℜ}. 

Example 5:

Solve $\sqrt{4x-12}\ge x-3$√4x−12≥x−3

We first recognise the requirement to have $4x-12>=0$4x−12>=0. This means that $4x>=12$4x>=12 or that $x>=3$x>=3.

Then we solve the inequality as follows:

$\sqrt{4x-12}$√4x−12	$>=$>=	$x-3$x−3
$4x-12$4x−12	$>=$>=	$\left(x-3\right)^2$(x−3)2
$4x-12$4x−12	$>=$>=	$x^2-6x+9$x2−6x+9
$0$0	$>=$>=	$x^2-10x+21$x2−10x+21
$0$0	$>=$>=	$\left(x-3\right)\left(x-7\right)$(x−3)(x−7)
$\left(x-3\right)\left(x-7\right)$(x−3)(x−7)	$<=$<=	$0$0

Various values of $x$x substituted into this last inequality reveal that the product $\left(x-3\right)\left(x-7\right)$(x−3)(x−7) remains negative or zero whenever $x$x is in the interval $3\le x\le7$3≤x≤7. This interval is entirely within the interval $x>=3$x>=3 deduced from our first observation, so we can correctly state that the solution is given by $\left\{x:3\le x\le7,x\in\Re\right\}${x:3≤x≤7,x∈ℜ}.  

Graphical techniques 

We will redo Example $5$5 using a graphing technique. Essentially, if we are confronted with the inequality $f\left(x\right)\ge g\left(x\right)$f(x)≥g(x), irrespective of what $f\left(x\right)$f(x) and $g\left(x\right)$g(x) are, we simply sketch both functions on the same graph and observe the interval (if it exists) where $f\left(x\right)\ge g\left(x\right)$f(x)≥g(x). Of course, it requires a good understanding of the basic functions, and their sketches.

Example 6:

Graphically solve $\sqrt{4x-12}\ge x-3$√4x−12≥x−3

Here we sketch $f\left(x\right)=\sqrt{4x-12}$f(x)=√4x−12 and $g\left(x\right)=x-3$g(x)=x−3 and see where $f\left(x\right)\ge g\left(x\right)$f(x)≥g(x). 

The solution is clear enough, but it is often the case that the intersection points are not obvious. To get around this we would attempt to solve the equation $f\left(x\right)=g\left(x\right)$f(x)=g(x).

 

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4