Quadratic Inequalities: Problems with Solutions

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: $x\in \left( -\infty ,-5\right) \cup \left( 3,+\infty \right)$

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: $x\in \left[ -1,3\right]$

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] $

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) $

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) $

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) $

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) $

Solution:
Let's solve the equation
$9x^{2}+30x+25=0$

$x=\frac{-30\pm \sqrt{900-4\left( 9\right) \left( 25\right) }}{18}=\frac{-30\pm \sqrt{900-900}}{18}=-\frac{5}{3}$ then
$9x^{2}+30x+25=9\left(x+\frac{5}{3}\right)^{2}$

But $\left( x+\frac{5}{3}\right) ^{2}>0$ for all $x\neq -\frac{5}{3}$
Answer: $x\in \left( -\infty ,-\frac{5}{3}\right) \cup \left( -\frac{5}{3},+\infty \right)$

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 $x\in \mathbb{R}$

Answer: $x\in \varnothing$

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 $x+3=0\Longrightarrow x=-3$ is the solution.

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: $x\in \left( -3,-2\right) \cup \left( -1,\infty \right)$

Solution:
$-2\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) \leq 0$

Find the roots of the equation

$\left( x-1\right) \left( x+\frac{1}{2}\right) \left( x-3\right) =0$

$x=-\frac{1}{2}\qquad x=1\qquad x=3$

We 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) $

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 sign table 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] $

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 sign 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) $

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 sign 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) $

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

$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x-3\right) & \left( -\right) & -6 & \left( -\right) & 0 & \left( +\right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 6 & \left( +\right) \\ \left( x-3\right) \left( x+3\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$

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.

Solution:
$d=\frac{1}{2}\left( n-1\right) n-n$

$d>35\qquad \frac{1}{2}\left(n-1\right) n-n>35$

$n^{2}-n-2n>70\qquad n^{2}-3n>70\qquad n^{2}-3n-70>0$

$\left( n-10\right) \left( n+7\right) >0$

$\begin{array}{cccccc} & \left( -\infty ,-7\right) & -7 & \left( -7,10\right) & 10 & \left( 10,+\infty \right) \\ \left( n-10\right) & \left( -\right) & -17 & \left( -\right) & 0 & \left( +\right) \\ \left( n+7\right) & \left( -\right) & 0 & \left( +\right) & 17 & \left( +\right) \\ \left( n-10\right) \left( n+7\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$

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 $n>10$.

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 $n\geq 1$ we get that $n\in \left[ 1,100\right) $
so $1\leq n< 100$

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