Mathematics » Equations and Inequalities » Quadratic Inequalities

Quadratic Inequalities

Quadratic Inequalities

Quadratic inequalities can be of the following forms:

\begin{align*} ax^2 + bx + c > 0 \\ ax^2 + bx + c \ge 0 \\ ax^2 + bx + c < 0 \\ ax^2 + bx + c \le 0\end{align*}

To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the \(x\)-axis. An inequality can therefore be solved graphically using a graph or algebraically using a table of signs to determine where the function is positive and negative.

Example

Question

Solve for \(x\): \({x}^{2}-5x+6\ge 0\)

Factorise the quadratic

\[(x-3)(x-2) \geq 0\]

Determine the critical values of \(x\)

From the factorised quadratic we see that the values for which the inequality is equal to zero are \(x=3\) and \(x=2\). These are called the critical values of the inequality and they are used to complete a table of signs.

Complete a table of signs

We must determine where each factor of the inequality is positive and negative on the number line:

  • to the left (in the negative direction) of the critical value
  • equal to the critical value
  • to the right (in the positive direction) of the critical value

In the final row of the table we determine where the inequality is positive and negative by finding the product of the factors and their respective signs.

Critical values \(x=2\) \(x=3\) 
\(x-3\)\(-\)\(-\)\(-\)\(\text{0}\)\(+\)
\(x-2\)\(-\)\(\text{0}\)\(+\)\(+\)\(+\)
\(f(x)=(x-3)(x-2)\)\(+\)\(\text{0}\)\(-\)\(\text{0}\)\(+\)

From the table we see that \(f(x)\) is greater than or equal to zero for \(x\le 2\) or \(x\ge 3\).

A rough sketch of the graph

The graph below does not form part of the answer and is included for illustration purposes only. A graph of the quadratic helps us determine the answer to the inequality. We can find the answer graphically by seeing where the graph lies above or below the \(x\)-axis.

  • From the standard form, \(x^2 – 5x + 6\), \(a>0\) and therefore the graph is a “smile” and has a minimum turning point.
  • From the factorised form, \((x-3)(x-2)\), we know the \(x\)-intercepts are \((2;0)\) and \((3;0)\).

94a4933defc7ed765ffb6fe89f3df66c.png

The graph is above or on the \(x\)-axis for \(x \le 2\) or \(x \ge 3\).

Write the final answer and represent on a number line

\[x^2 – 5x + 6 \geq 0 \text{ for } x \leq 2 \text{ or } x \geq 3\]

3743b86d44b46fd8f1330c2e6f13cb14.png

Example

Question

Solve for \(x\): \(4{x}^{2}-4x+1\le 0\)

Factorise the quadratic

\begin{align*} (2x-1)(2x-1) & \leq 0 \\ (2x-1)^2 & \leq 0 \end{align*}

Determine the critical values of \(x\)

From the factorised quadratic we see that the value for which the inequality is equal to zero is \(x=\cfrac{1}{2}\). We know that \(a^2 > 0\) for any real number \(a, a \ne 0\), so then \((2x-1)^2\) will never be negative.

A rough sketch of the graph

The graph below does not form part of the answer and is included for illustration purposes only.

  • From the standard form, \(4x^2 – 4x + 1\), \(a > 0\) and therefore the graph is a “smile” and has a minimum turning point.
  • From the factorised form, \((2x-1)(2x-1)\), we know there is only one \(x\)-intercept at \(\left(\cfrac{1}{2};0)\).

532e07c228f76bcae2cf18b3c025db2e.png

Notice that no part of the graph lies below the \(x\)-axis.

Write the final answer and represent on a number line

\[4x^2 – 4x + 1 \leq 0 \text{ for } x = \cfrac{1}{2}\]

adc4f7f25425506d7d6128b2fd2a3260.png

Example

Question

Solve for \(x\): \(-{x}^{2}-3x+5>0\)

Examine the form of the inequality

Notice that the coefficient of the \(x^2\) term is \(-\text{1}\). Remember that if we multiply or divide an inequality by a negative number, then the inequality sign changes direction. So we can write the same inequality in different ways and still get the same answer, as shown below.

\[-x^2-3x+5 > 0\]

Multiply by \(-\text{1}\) and change direction of the inequality sign \[x^2+3x-5 < 0\]

4eaf24c673f9c70d9c54bee7f13f339c.pnge30395e6bf4da333439037e522ecd080.png

From this rough sketch, we can see that both inequalities give the same solution; the values of \(x\) that lie between the two \(x\)-intercepts.

Factorise the quadratic

We notice that \(-{x}^{2}-3x+5>0\) cannot be easily factorised. So we let \(-x^2-3x+5=0\) and use the quadratic formula to determine the roots of the equation.

\begin{align*} -x^2 – 3x + 5 &= 0 \\ x^2 + 3x – 5 &= 0\end{align*}\begin{align*} \therefore x &= \cfrac{-3±\sqrt{{(3)}^{2}-4(1)(-5)}}{2(1)} \\ &= \cfrac{-3±\sqrt{29}}{2} \\ x_1 &= \cfrac{-3-\sqrt{29}}{2} \approx -\text{4.2} \\ x_2 &= \cfrac{-3+\sqrt{29}}{2} \approx \text{1.2}\end{align*}

Therefore we can write, correct to one decimal place, \begin{align*} x^2+3x-5 &< 0 \\ \text{as } (x-\text{1.2}\right)(x+\text{4.2}) &< 0 \end{align*}

Determine the critical values of \(x\)

From the factorised quadratic we see that the critical values are \(x=\text{1.2}\) and \(x=-\text{4.2}\).

Complete a table of signs

Critical values \(x=-\text{4.2}\) \(x=\text{1.2}\) 
\(x+\text{4.2}\)\(-\)\(\text{0}\)\(+\)\(+\)\(+\)
\(x-\text{1.2}\)\(-\)\(-\)\(-\)\(\text{0}\)\(+\)
\(f(x)=(x+\text{4.2})(x-\text{1.2})\)\(+\)\(\text{0}\)\(-\)\(\text{0}\)\(+\)

From the table we see that the function is negative for \(-\text{4.2} < x < \text{1.2}\).

A sketch of the graph

  • From the standard form, \(x^2+3x-5\), \(a > 0\) and therefore the graph is a “smile” and has a minimum turning point.
  • From the factorised form, \((x-\text{1.2})(x+\text{4.2})\), we know the \(x\)-intercepts are \((-\text{4.2};0)\) and \((\text{1.2};0)\).

114ca7eeaa0f73f74aa8dd7444071914.png

From the graph we see that the function lies below the \(x\)-axis between \(-\text{4.2}\) and \(\text{1.2}\).

Write the final answer and represent on a number line

\[x^2 + 3x – 5 < 0 \text{ for } -\text{4.2} < x < \text{1.2}\]

b5d0d516d69f0252584002cc813c7c41.png

Important: When working with an inequality in which the variable is in the denominator, a different approach is needed. Always remember to check for restrictions.

Example

Question

Solve for \(x\):

  1. \(\dfrac{2}{x+3} = \dfrac{1}{x-3}\), \(x \ne \pm 3\)

  2. \(\dfrac{2}{x+3} \le \dfrac{1}{x-3}\), \(x \ne \pm 3\)

Solving the equation

To solve this equation we multiply both sides of the equation by \((x+3)(x-3)\) and simplfy: \begin{align*} \cfrac{2}{x+3} \times (x+3)(x-3) &= \cfrac{1}{x-3} \times (x+3)(x-3)\\ 2(x-3) &= x+3\\ 2x-6 &= x+3\\ x &= 9 \end{align*}

Solving the inequality

It is very important to recognise that we cannot use the same method as above to solve the inequality. If we multiply or divide an inequality by a negative number, then the inequality sign changes direction. We must rather simplify the inequality to have a lowest common denominator and use a table of signs to determine the values that satisfy the inequality.

Subtract \(\dfrac{1}{x-3}\) from both sides of the inequality

\[\cfrac{2}{x+3} – \cfrac{1}{x-3} \le 0\]

Determine the lowest common denominator and simplify the fraction

\begin{align*} \cfrac{2(x-3) – (x+3)}{(x+3)(x-3)} &\le 0 \\ \cfrac{x-9}{(x+3)(x-3)} &\le 0 \end{align*}

Keep the denominator because it affects the final answer.

Determine the critical values of \(x\)

From the factorised inequality we see that the critical values are \(x=-3\), \(x=3\) and \(x=9\).

Complete a table of signs

Critical values \(x=-3\) \(x=3\) \(x=9\) 
\(x+3\)\(-\)undef\(+\)\(+\)\(+\)\(+\)\(+\)
\(x-3\)\(-\)\(-\)\(-\)undef\(+\)\(+\)\(+\)
\(x-9\)\(-\)\(-\)\(-\)\(-\)\(-\)\(\text{0}\)\(+\)
\(f(x)=\dfrac{x-9}{(x+3)(x-3)}\)\(-\)undef\(+\)undef\(-\)\(\text{0}\)\(+\)

From the table we see that the function is less than or equal to zero for \(x < -3\) or \(3 < x \leq 9\). We do not include \(x = -3\) or \(x = 3\) in the solution because of the restrictions on the denominator.

Write the final answer and represent on a number line

\[x < -3 \quad\text{or}\quad 3 < x \le 9\]

3cc26d8d3933066b305f7f23c3d22f82.png

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