Solving a Second Degree Equation or Inequality

A second degree equation with one variable is an equation reducible to the form |ax^2+bx+c=0|, where |x| is the variable, |a \in \mathbb{R}^*| and |b,c \in \mathbb{R}|.

The main steps in the problem solving process are the following.

1. Change the quadratic equation to the form |ax^2+bx+c=0|, if it is not already the case.

2. Evaluate the discriminant |b^2-4ac| and check whether it is worth continuing. If |b^2-4ac<0|, there is no solution.

3. If |b^2-4ac \geq 0|, check if it is possible to factor easily.

4. Apply the rule of the zero product to find the values of |x| sought.

5. Give the solution set.

For example, consider the equation |2x^2+9x+5=-4.|

1. Change the equation to the form |ax^2+bx+c=0| by adding |4| to each side of the equality. ||2x^2+9x+5=-4\ \rightarrow\ 2x^2+9x+9=0||

2. Evaluate the discriminant |b^2-4ac|, where |a=2, b=9| and |c=9.| ||b^2-4ac=(9)^2-4(2)(9) = 9||
   We can continue since the discriminant is positive.

3. |2x^2+9x+9| can be factored using the product-sum method. ||2x^2+9x+9=0\ \rightarrow\ (x+3)(2x+3)=0||

4. Apply the zero product rule. ||x+3 = 0\ \Rightarrow\ x=-3|| ||\text{or}|| ||2x+3 = 0 \Rightarrow\ x=-\frac{3}{2}||

5. Therefore, the starting equation’s two solutions are |-3| and |-\dfrac{3}{2}.| 

The main steps in the problem solving process are the following.

1. Change the quadratic equation to be of the form |ax^2+bx+c=0,| if it is not already the case.

2. Evaluate the discriminant |b^2-4ac| and check whether it is worth continuing. If |b^2-4ac<0,| the equation has no solution.

3. Complete the square.

4. Apply the zero product rule to find the values of |x| sought.

5. Give the solution set.

For example, consider the equation |2x^2=-3x+5.|

1. Change the equation to be of the form |ax^2+bx+c=0| by adding |3x| and subtracting |5| on both sides of the equality. ||2x^2=-3x+5\ \rightarrow\ 2x^2+3x-5=0||

2. Evaluate the discriminant |b^2-4ac| where |a=2, b=3| and |c=-5.| ||b^2-4ac = 3^2-4(2)(-5) = 49||
   We can continue since the discriminant is non-zero.

3. |2x^2+3x-5| can be factored by completing the square. ||\begin{align} 2x^2+3x-5=0\ &\rightarrow\ 2\left(x+\frac{10}{4}\right)\left(x-\frac{4}{4}\right)=0\\ &\rightarrow\ \left(x+\frac{5}{2}\right)(x-1)=0 \end{align}||

4. Apply the zero product rule. ||x + \dfrac{5}{2} = 0\ \Rightarrow\ x = -\dfrac{5}{2}|| ||\text{or}|| ||x-1 = 0\ \Rightarrow\ x =1||

5. The solution set is |\left\lbrace -\dfrac{5}{2}, 1 \right\rbrace.| 

The main steps in the problem solving process are the following.

1. Change the quadratic equation to be of the form |ax^2+bx+c=0|, if it is not already the case.

2. Evaluate the discriminant |b^2-4ac| to check whether it is worth continuing.

3. If |b^2-4ac<0|, the equation has no solution.

4. Use the quadratic formula |\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.|

5. Give the solution set.

For example, consider the equation |x^2-4x-20=0.|

1. The equation is already in the general form.

2. Calculate the discriminant |b^2-4ac| where |a=1,b=-4| and |c=-20|  
   
   |b^2-4ac= (-4)^2 - 4(1)(-20) = 96|
   Therefore, we can continue.

3. Use the quadratic formula
   |x_{1,2} = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{--4 \pm \sqrt{(-4)^2-4(1)(-20)}}{2 \times 1}|  
   |x_{1,2} = \dfrac{4 \pm \sqrt{96}}{2}|
   Split the equation into two. |x_1 = \dfrac{4 + \sqrt{96}}{2} \approx 6{.}9|
   or
   |x_2 = \dfrac{4 - \sqrt{96}}{2} \approx -2{.}9|

4. The solutions are |-2{.}9| and |6{.}9.| 

A second degree inequality with one variable is an inequality which can be reduced to one of the forms below.

|ax^2+bx+c>0|
|ax^2+bx+c<0|
|ax^2+bx+c \geq 0|
|ax^2+bx + c \leq 0|
|a(x-h)^2+k >0|
|a(x-h)^2+k<0|
|a(x-h)^2+k \geq 0|
|a(x-h)^2+k \leq 0|

where |x| is the variable, |a \in \mathbb{R}^*| and |b,c \in \mathbb{R}.|

1. Represent the inequality with a graph by indicating the solution set.

2. Determine the points of intersection’s coordinates by solving the system of equations.

3. Determine the interval of the values of |x| that respect the inequality.

For example, consider the equation |-3x^2-5x+7 \geq 2x+1.|

1. Represent the inequality with a graph by indicating the solution set

In this case, we are interested in the section of the quadratic function that is greater than or equal to the linear function. Due to the inequality sign, the intersection points are represented by solid points. 

2. Determine the points of intersection’s coordinates by solving the system of equations

||\begin{align}
-3x^2-5x + 7 \geq 2x + 1 \Rightarrow & -3x^2 - 5x + 7 && = && 2x+1 \\
& -3x^2 - 7x + 6 && = && 0 \\
& -3x^2 -9x + 2x + 6 && = && 0 \\
& -3x (x + 3) + 2 (x+3) && = && 0 \\
& (x+3) (-3x + 2) && = && 0 \\
& (x+3) = 0 && \text{OR} && -3x + 2 = 0 \\
& x = -3 && \text{OR} && x = \frac{2}{3} \end{align}||
Note: the product-sum factoring method was used and it is not necessary to find the values in |y| of each coordinate in the example. 

3. Determine the interval of |x| values that respect the inequality

According to the previous graph, it is determined that the values of |x| must be in the interval |[-3, \frac{2}{3}]|.

For example, consider the equation |2(x-1)^2+3<0.|

1. Represent the inequality with a graph by indicating the solution set

We can immediately conclude the solution set is empty, i.e., there is no value of |x| that respects the inequality.

Note: however, if the initial equation had been |2(x-1)^2+3>0| instead, the solution set would have been the entire set of real numbers, denoted by |\mathbb{R}|.

1. Transform the inequality so the right hand side equals to zero.

2. Factor the trinomial.

3. Determine the value of |x| which cancels out each factor.

4. Build a four row and six column sign table.
   - In the first column, there is |x| , the two binomials, and the complete algebraic expression.
   - In boxes three and five, there are the values of |x| found in step three in ascending order.
   - Place |0| under the values of |x| which cancel out the corresponding factor.

5. Fill rows two and three with |+| and |-| signs according to the value of the binomial.

6. Complete the last row by multiplying rows two and three.

7. Give the solution set.

For example, consider the equation |2x^2-10 >-x|.

1. Transform the inequality so that the right hand side is zero
   
   |2x^2 -10 > -x \ \ \Rightarrow\ \  2x^2+x-10>0|

2. Factor in the trinomial
   
   |\begin{align} 2x^2+x-10 &>0\\ 2x^2+5x-4x-10 &>0\\ x(2x+5)-2(2x+5) &>0\\ (2x+5)(x-2)&>0 \end{align}|

3. Determine the value of |x| which cancels out each factor
   
   |\begin{align}2x+5=0 \ \ &\Rightarrow\ \  x_1=-\dfrac{5}{2}\\x-2=0 \ \ &\Rightarrow\ \  x_2=2\end{align}|

4. Build a table of signs

   ​|x|	​	​|-\dfrac{5}{2}|	​	​|2|	​
   ​|2x+5|		​|0|
   ​|x-2|		​		|0|​	​
   ​​|(2x+5)(x-2)|		​			​

5. Fill rows two and three with |+| and |-| signs according to the value of the binomial

   For the expression |2x+5| , since the coefficient in front of the |x| is positive, it is an increasing line. Thus, the value of the expression is negative before its zero |(x_1=-\frac{5}{2})| and positive afterwards. Therefore, in the table, place the |-| sign in the box located before |-\frac{5}{2}| and the sign |+| in the boxes after.

   Regarding the expression |x-2| , the slope is also positive. Thus, we place the |-| sign in the boxes preceding its zero |(x_2=2)| and the |+| sign after.

   The result is the following table.

   ​|x|	​	​|-\dfrac{5}{2}|	​	​|2|	​
   ​|2x+5|	|-|	​|0|	|+|	|+|	|+|
   ​|x-2|	|-|	|-|	|-|	|0|​	|+|
   ​|(2x+5)(x-2)|		​			​

   There is always a change of sign on either side of a zero. The signs could also have been determined by calculating the value of the expression with a value |x| as desired (number |0| is often a good choice).

6. The last row is obtained by multiplying rows two and three
   
   The law of signs says that the product of two opposite signs gives a |-,| while the product of two identical signs gives a |+.| Moreover, if we multiply anything by |0,| we necessarily get |0.|
   
   The table is now complete.

   ​|x|	​	​|-\dfrac{5}{2}|	​	​|2|	​
   ​|2x+5|	|-|	​|0|	|+|	|+|	|+|
   ​|x-2|	|-|	​|-|	|-|	|0|​	|+|
   ​|(2x+5)(x-2)|	|+|	|0|​	|-|	|0|	​|+|

    

7. Give the solution set.
   
   According to the inequation obtained in step two, we must give the interval of |x| ensuring that |(2x+5)(x-2)| is positive. From the last row of the table, it can be determined that the solution set is |\left]-\infty,-\dfrac{5}{2}\right[\ \cup\ \bigg]2,+\infty\bigg[.|
   
   The boundaries of the intervals are excluded since the sign of inequality is |>.|