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Practice
Quadratic Inequalities
Quadratic Inequalities: Problems with Solutions
By
Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
What is the solution to the inequality?
$x^{2}+2x-15>0$
$x\in \left( -\infty ,-3\right) \cup \left( 5,+\infty \right) $
$x\in \left( -\infty ,5\right) \cup \left( 3,+\infty \right) $
$x\in \left( -\infty ,-5\right) \cup \left( 3,+\infty \right) $
$x\in \left( -5,3\right) $
Solution:
$x^{2}+2x-15>0$
Factor the left side $\left( x+5\right) \left(x-3\right) >0$
The zeros are $x=-5\qquad x=3$
$\begin{array}{cccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x+5\right) & \left( -\right) & 0 & \left( +\right) & 8 & \left( +\right) \\ \left( x-3\right) & \left( -\right) & -8 & \left( -\right) & 0 & \left( +\right) \\ \left( x+5\right) \left( x-3\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Solution: $x\in \left( -\infty ,-5\right) \cup \left( 3,+\infty \right)$
Problem 2
Solve the inequality by factoring the expression on the left side.
$x^{2}-2x-3\leq 0$
$x\in \left[ -1,3\right] $
$x\in \left( -1,3\right) $
$x\in \left( -\infty ,-1\right) $
$x\in \left( 3,+\infty \right) $
Solution:
$x^{2}-2x-3\leq 0$
$\left( x+1\right) \left(x-3\right) \leq 0$
The roots are $x=-1\qquad x=3$
$\begin{array}{cccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x+1\right) & \left( -\right) & 0 & \left( +\right) & 4 & \left( +\right) \\ \left( x-3\right) & \left( -\right) & -4 & \left( -\right) & 0 & \left( +\right) \\ \left( x+1\right) \left( x-3\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left[ -1,3\right]$
Problem 3
$3x^{2}-x-2\leq 0$
$x\in \left[ -2,3\right] $
$x\in (-\infty ,1]$
$x\in \lbrack 1,\infty )$
$x\in \left[ -\frac{2}{3},1\right] $
Solution:
$3x^{2}-x-2\leq 0$
The roots of the quadratic equation are:
$x=\frac{1\pm \sqrt{1-4\left( 3\right) \left( -2\right) }}{6}=\frac{1\pm \sqrt{25}}{6}$
$x=\frac{1\pm 5}{6}$
$x=1\qquad x=-\frac{2}{3}$
then
$3x^{2}-x-2=3\left( x-1\right) \left( x+\frac{2}{3}\right) \leq 0$
$\begin{array}{cccccc} & \left( -\infty ,-\frac{2}{3}\right) & -\frac{2}{3} & \left( -\frac{2}{3},1\right) & 1 & \left( 1,+\infty \right) \\ \left( x-1\right) & \left( -\right) & -\frac{5}{3} & \left( -\right) & 0 & \left( +\right) \\ \left( x+\frac{2}{3}\right) & \left( -\right) & 0 & \left( +\right) & \frac{1}{3} & \left( +\right) \\ 3\left( x-1\right) \left( x+\frac{2}{3}\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left[ -\frac{2}{3},1\right] $
Problem 4
Solve the inequality by factoring the expression on the left side.
$x^{2}-8x+12<0$
$x\in \left( -\infty ,2\right) \cup \left( 6,+\infty \right) $
$x\in \left( 6,+\infty \right) $
$x\in \left( 2,6\right) $
$x\in \left( -\infty ,2\right) $
Solution:
$x^{2}-8x+12=\left( x-2\right) \left( x-6\right) <0$
$x=2$
$x=6$
$\begin{array}{cccccc} & \left( -\infty ,2\right) & 2 & \left( 2,6\right) & 6 & \left( 6,+\infty \right) \\ \left( x-2\right) & \left( -\right) & 0 & \left( +\right) & 4 & \left( +\right) \\ \left( x-6\right) & \left( -\right) & -4 & \left( -\right) & 0 & \left( +\right) \\ \left( x-2\right) \left( x-6\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left( 2,6\right) $
Problem 5
Solve the inequality by factoring the expression on the left side.
$x^{2}-5x\geq 0$
$x\in \left[ -\infty ,0\right] \cup \left[ 5,+\infty \right] $
$x\in \left[ -\infty ,0\right] $
$x\in \left[ 5,+\infty \right] $
$x\in \left[ 0,5\right] $
Solution:
We need to solve $x^{2}-5x=0$
$x^{2}-5x=x\left( x-5\right)$
So the roots of the equation are:
$x=0\qquad x=5$
$\begin{array}{cccccc} & \left( -\infty ,0\right) & 0 & \left( 0,5\right) & 5 & \left( 5,+\infty \right) \\ x & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) \\ \left( x-5\right) & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) \\ x\left( x-5\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left[ -\infty ,0\right] \cup \left[ 5,+\infty \right] $
Problem 6
$3x^{2}-27<0$
$x\in \left( -\infty ,-3\right) \cup \left( 3,+\infty \right) $
$x\in \left( 3,+\infty \right) $
$x\in \left( -3,3\right) $
$x\in \left( -\infty ,-3\right) $
Solution:
We need to find the zeros to $3x^{2}-27=0$
$3x^{2}-27=3\left( x-3\right) \left( x+3\right)$
So the roots are:
$x=-3\qquad x=3$
Let's look at $3x^{2}-27<0$
$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x-3\right) & \left( -\right) & -6 & \left( -\right) & 0 & \left( +\right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 6 & \left( +\right) \\ 3\left( x-3\right) \left( x+3\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left( -3,3\right) $
Problem 7
$9x>2x^{2}-18$
$x\in \left( -\infty ,-\frac{3}{2}\right) $
$x\in \left( 6,+\infty \right) $
$x\in \left( -\infty ,-\frac{3}{2}\right) \cup \left( 6,+\infty \right) $
$x\in \left( -\frac{3}{2},6\right) $
Solution:
$9x>2x^{2}-18\Longrightarrow 2x^{2}-9x-18<0$
The roots are:
$x=\frac{9\pm \sqrt{81-4\left( 2\right) \left( -18\right) }}{4}=\frac{9\pm \sqrt{225}}{4}=$ $\frac{9\pm 15}{4}$
$x=6\qquad x=-\frac{3}{2}$
We can write the left side as product of two linear factors:
$2\left( x-6\right) \left( x+\frac{3}{2}\right)$
$\begin{array}{cccccc} & \left( -\infty ,-\frac{3}{2}\right) & -\frac{3}{2} & \left( -\frac{3}{2}% ,6\right) & 6 & \left( 6,+\infty \right) \\ \left( x-6\right) & \left( -\right) & -\frac{15}{2} & \left( -\right) & 0 & \left( +\right) \\ \left( x+\frac{3}{2}\right) & \left( -\right) & 0 & \left( +\right) & \frac{% 15}{2} & \left( +\right) \\ 2\left( x-6\right) \left( x+\frac{3}{2}\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left( -\frac{3}{2},6\right) $
Problem 8
$9x^{2}+30x>-25$
$x\in \left( 0,\frac{15}{9}\right) $
$x\in \left( -\infty ,\frac{15}{9}\right) \cup \left( \frac{15}{9},\infty \right)$
$x\in \left( \frac{15}{9},\infty \right) $
$x\in \left( -\infty ,\frac{15}{9}\right) $
Solution:
Let's solve the equation
$9x^{2}+30x+25=0$
$x=\frac{-30\pm \sqrt{900-4\left( 9\right) \left( 25\right) }}{18}=\frac{-30\pm \sqrt{900-900}}{18}=-\frac{15}{9}$ then
$9x^{2}+30x+25=9\left(x+\frac{15}{9}\right)^{2}$
But $\left( x+\frac{15}{9}\right) ^{2}>0$
For all $x\neq \frac{15}{9}$
Answer: $x\in \left( -\infty ,\frac{15}{9}\right) \cup \left( \frac{15}{9},\infty \right)$
Problem 9
$4x^{2}-4x+1<0$
$x\in \left( -\infty ,1\right) $
$x\in \left( -\infty ,-4\right) $
$x\in \left( -\infty ,-\frac{1}{2}\right) $
$x\in \phi $
Solution:
The roots of the equation
$4x^{2}-4x+1=0$
are $x=\frac{4\pm \sqrt{16-4(4)1}}{8}=\frac{1}{2}$
$4x^{2}-4x+1=4\left( x-\frac{1}{2}\right)^{2}<0$
The quadratic inequality $4x^{2}-4x+1<0$ has no real solutions since
$\left( x-\frac{1}{2}\right) ^{2}\geq 0$ for all $x\in R$
Answer: $x\in \phi$
Problem 10
Solve the inequality by factoring the expression on the left side.
$x^{2}+6x\leq -9$
$x\in \left( -3,3\right) $
$x\in \left( -\infty ,3\right) $
$x=-3$
$x\in \phi $
Solution:
$x^{2}+6x\leq -9\Longrightarrow x^{2}+6x+9\leq 0$
Then $\left( x+3\right)^{2}\leq 0$
Since $\left( x+3\right) ^{2}=0$ only if $x+3\Longrightarrow x=-3$ is the solution.
Problem 11
Solve the following inequiality
$-\left( x+1\right)\left( x+2\right) \left( x+3\right) < 0$
$x\in \left( -3,-2\right) \cup \left( -1,\infty \right) $
$x\in \left( -3,-2\right) $
$x\in \left( -1,\infty \right) $
$x\in \left( -2,-1\right) $
Solution:
Find the roots of the equation
$\left( x+1\right) \left( x+2\right) \left( x+3\right) =0$
$x=-1\qquad x=-2\qquad x=-3$
We obtain the following table:
$\begin{array}{cccccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,-2\right) & -2 & \left( -2,-1\right) & -1 & \left( -1,\infty \right) \\ \left( x+1\right) & \left( -\right) & -2 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ \left( x+2\right) & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 2 & (+) \\ \left( x+1\right) \left( x+2\right) \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
and since $-\left( x+1\right) \left( x+2\right) \left( x+3\right) <0\Longrightarrow \left( x+1\right) \left( x+2\right) \left( x+3\right) >0$
Answer: $x\in \left( -3,-2\right) \cup \left( -1,\infty \right)$
Problem 12
Solve the following inequiality
$-2\left( x-1\right)\left( x+\frac{1}{2}\right) \left( x-3\right) \leq 0$
$x\in \left[ -\frac{1}{2},1\right] $
$x\in \left[ -\frac{1}{2},1\right] \cup \left[ 3,\infty \right] $
$x\in \left[ 3,\infty \right] $
$x\in \Phi $
Solution:
$-2\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) \leq 0$
Find the roots of the equation
$\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) =0$
$x=-\frac{1}{2}\qquad x=1\qquad x=3$
Obtain the following table
$\begin{array}{cccccccc} & \left( -\infty ,-\frac{1}{2}\right) & -\frac{1}{2} & \left( -\frac{1}{2},1\right) & 1 & \left( 1,3\right) & 3 & \left( 3,\infty \right) \\ \left( x-1\right) & \left( -\right) & -\frac{3}{2} & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ \left( x+\frac{1}{2}\right) & \left( -\right) & 0 & \left( +\right) & \frac{3}{2} & \left( +\right) & \frac{7}{2} & \left( +\right) \\ \left( x-3\right) & \left( -\right) & -\frac{7}{2} & \left( -\right) & -2 & \left( -\right) & 0 & (+) \\ \left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) & \left( -\right) & 0 & \left( +\right) & 0 & \left( +\right) & 0 & \left( +\right) \end{array}$
Since $-2\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) \leq 0$ when $\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) \geq 0$
Answer: $x\in \left[ -\frac{1}{2},1\right] \cup \left[ 3,\infty \right] $
Problem 13
$\left( x^{2}-1\right) \left(x^{2}-4\right) \leq 0$
$x\in \left[ -2,-1\right] \cup \left[ 1,2\right] $
$x\in \left[ -2,-1\right] $
$x\in \left[ 1,2\right] $
$x\in \left[ -1,1\right] $
Solution:
$\left( x^{2}-1\right) \left( x^{2}-4\right) =0\Longrightarrow \left( x-1\right) \left( x+1\right) \left( x-2\right) \left( x+2\right) =0$
The roots are $x=1\qquad x=-1\qquad x=2\qquad x=-2$
Then the table of monotony intervals is
$\begin{array}{cccccccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,2\right) & 2 & \left( 2,+\infty \right) \\ \left( x-1\right) & \left( -\right) & -3 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ \left( x+1\right) & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 3 & \left( +\right) \\ \left( x-2\right) & \left( -\right) & -4 & \left( -\right) & -3 & \left( -\right) & -1 & (-) & 0 & \left( +\right) \\ \left( x+2\right) & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 3 & \left( +\right) & 4 & \left( +\right) \\ \left( x-1\right) \left( x+1\right) \left( x-2\right) \left( x+2\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
So $\left( x^{2}-1\right) \left( x^{2}-4\right) \leq 0$ when $x\in \left[ -2,-1\right] \cup \left[ 1,2\right] $
Problem 14
$\left( x-1\right)^{2}\left( x+3\right) \left( x+5\right) >0$
$x\in \left( -\infty ,-5\right) \cup \left( -3,1\right) $
$x\in \left( -\infty ,-5\right) \cup \left( 1,+\infty \right) $
$x\in \left( -3,1\right) \cup \left( 1,+\infty \right) $
$x\in \left( -\infty ,-5\right) \cup \left( -3,1\right) \cup \left( 1,+\infty \right) $
Solution:
We must know the zeros of $\left( x-1\right) ^{2}\left( x+3\right) \left(x+5\right) =0$
They are $x=1\qquad x=-3\qquad x=-5$
The monotony table is
$\begin{array}{cccccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,-3\right) & -3 & \left( -3,1\right) & 1 & \left( 1,+\infty \right) \\ \left( x-1\right) ^{2} & \left( +\right) & 36 & \left( +\right) & 16 & \left( +\right) & 0 & \left( +\right) \\ \left( x+3\right) & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 4 & \left( +\right) \\ \left( x+5\right) & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 6 & \left( +\right) \\ \left( x-1\right) ^{2}\left( x+3\right) \left( x+5\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left( -\infty ,-5\right) \cup \left( -3,1\right) \cup \left( 1,+\infty \right) $
Problem 15
$\left( x+3\right) ^{2}\left( x+4\right) \left( x-5\right)^{3}>0$
$x\in \left( -\infty ,-4\right) $
$x\in \left( -\infty ,-4\right) \cup \left( 5,+\infty \right) $
$x\in \left( -4,5\right) $
$x\in \left( 5,+\infty \right) $
Solution:
We must know the zeros of
$\left( x+3\right) ^{2}\left( x+4\right) \left( x-5\right) ^{3}=0$
They are $x=-3\qquad x=-4\qquad x=5$
The monotony table is
$\begin{array}{cccccccc} & \left( -\infty ,-4\right) & -4 & \left( -4,-3\right) & -3 & \left( -3,5\right) & 5 & \left( 5,+\infty \right) \\ \left( x+3\right) ^{2} & \left( +\right) & 1 & \left( +\right) & 0 & \left( +\right) & 64 & \left( +\right) \\ \left( x+4\right) & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 9 & \left( +\right) \\ \left( x-5\right) ^{3} & \left( -\right) & -729 & \left( -\right) & -512 & \left( -\right) & 0 & \left( +\right) \\ \left( x+3\right) ^{2}\left( x+4\right) \left( x-5\right) ^{3} & \left( +\right) & 0 & \left( -\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
Answer: $x\in \left( -\infty ,-4\right) \cup \left( 5,+\infty \right) $
Problem 16
If $7$ times the square of a positive number is reduced by $3$ and the result is greater than $60$, what can the number be?
$x>5$
$x>3$
$x>7$
$\mathbf{3< x < 7}$
Solution:
$7x^{2}-3>60\Longrightarrow 7x^{2}-63>0\Longrightarrow 7\left( x-3\right) \left( x+3\right) >0$
We obtain the following table
$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x-3\right) & \left( -\right) & -6 & \left( -\right) & 0 & \left( +\right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 6 & \left( +\right) \\ \left( x-8\right) \left( x+15\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
The solution to the inequality is $x\in \left( -\infty ,-3\right) \cup \left( 3,+\infty \right) $
But, since $x>0$ then $x\in \left( 3,+\infty \right)$
So $x>3$ is the answer to the problem.
Problem 17
The number of diagonals $d$ of an
n-sided
polygon is given by the formula $d=\frac{1}{2}\left( n-1\right) n-n$.
Which polygon has the number of diagonals greater than $35$?
$n>10\qquad n\in Z$
$n<10\qquad n\in Z$
$n>5\qquad n\in Z$
$5< n < 10\mathbf{\qquad }n\in Z$
Solution:
$d=\frac{1}{2}\left( n-1\right) n-n$
$d>35\qquad \frac{1}{2}\left(n-1\right) n-n>35$
$n^{2}-n-2n>70\qquad n^{2}-3n>70\qquad n^{2}-3n-70>0$
$\left( n-10\right) \left( n+7\right) >0$
$\begin{array}{cccccc} & \left( -\infty ,-7\right) & -7 & \left( -7,10\right) & 10 & \left( 10,+\infty \right) \\ \left( n-10\right) & \left( -\right) & -17 & \left( -\right) & 0 & \left( +\right) \\ \left( x+7\right) & \left( -\right) & 0 & \left( +\right) & 17 & \left( +\right) \\ \left( n-10\right) \left( n+7\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
The solution to the inequality is $n\in \left( -\infty ,-7\right) \cup \left( 10,+\infty \right)$
Since $n$ must be greater than $0$ then $n>10$.
Problem 18
The number $t$ of dots, ordered as shown below is given by the formula $t=\frac{n(n+1)}{2}$, where n is the number of rows.
Find the range of rows if the number of dots is less than $5050$.
$1\leq n<30$
$1\leq n<10$
$10\leq n<20$
$1\leq n<100$
Solution:
$t=\frac{n(n+1)}{2}\qquad t<5050$
$\frac{n(n+1)}{2}<5050$
$n^{2}+n<10100$
so $n^{2}+n-10100<0\Longrightarrow \left( n+101\right) \left( n-100\right) <0$
$\begin{array}{cccccc} & \left( -\infty ,-101\right) & -101 & \left( -101,100\right) & 100 & \left( 100,+\infty \right) \\ \left( n+101\right) & \left( -\right) & 0 & \left( +\right) & 201 & \left( +\right) \\ \left( n-100\right) & \left( -\right) & -201 & \left( -\right) & 0 & \left( +\right) \\ \left( n+101\right) \left( n-100\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$
The solution to the inequality is $n\in \left( -101,100\right)$
but, since $n>1$ we get that $n\in \left( 1,100\right) $
so $1\leq n< 100$
Problem 19
A rectangular garden should be twice as wide as it is long.
If the fenced area is greater than $98m^{2}$, what can we say about the width of the garden?
$x>1$
$x>24$
$x>14$
$x<98$
Solution:
$A=xy$
where $x=$ width and $y=$ length of the garden.
Since $x=2y\Longrightarrow y=\frac{1}{2}x$
so $A=xy=\frac{1}{2}x^{2}>98$
Then $x^{2}-196>0$
$\left( x-14\right) \left( x+14\right) >0$
We find the solution analyzing the following table
$\begin{array}{cccccc} & \left( -\infty ,-14\right) & -14 & \left( -14,14\right) & 14 & \left( 14,+\infty \right) \\ \left( x-14\right) & \left( -\right) & -28 & \left( -\right) & 0 & \left( +\right) \\ \left( x+14\right) & \left( -\right) & 0 & \left( +\right) & 28 & \left( +\right) \\ \left( x-14\right) \left( x+14\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array} $
The solution to the inequality is $x\in \left( -\infty ,-14\right) \cup \left( 14,+\infty \right) $
Since $x>0$ we obtain that $x\in \left( 14,+\infty \right) $ so
$x>14$ meters
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