Quadratic Equations Using Graphs and Tables

Key Concepts

• Solve quadratic equations using tables
• Solve quadratic equations using graphs
• Discriminant and intersection of graphs

Introduction 

A quadratic equation of the form ax² + bx + c = 0 can also be represented as y = ax² + bx + c.

The graph y = ax² + bx + c, is in the form of a parabola used to locate the solutions or roots of the quadratic equation. 

Solving Quadratic Equation graphically 

Here, the graph is y = ax2 + bx + c, which forms a parabola. 

The parabola intersects the x-axis at two points, which are called roots or solutions. 

The parabola intersects the y-axis at one point, which is constant c.   

To draw the graph given in the figure, firstly, we construct a table with x values and y values. 

We assume x values, and we find the y values to draw the graph.  

Two Solutions Quadratic Equation 

Example 1: 

Determine the solutions of the quadratic equation x²+2x−4=0 

by using graphs. Give answers to 1 decimal place where appropriate. 

Firstly, construct a table for y = x²+2x−4. 

Now, by the table, we have drawn the graph and plotted the points,  

The graph cuts the x-axis(y=0) at x = 1.23 and x = -3.23 

So, from the graph the solutions are x = 1.23 and x = -3.23 

Example 2: 

Determine the solutions of the quadratic equation x²−7x+12=0 

by using graphs. Give answers to 1 decimal place where appropriate. 

Firstly, construct a table for y = x²−7x+12. 

Now, by the table, we have drawn the graph and plotted the points,  

The graph cuts the x-axis(y=0) at x = 3 and x = 4. 

So, from the graph the solutions are x = 3 and x = 4. 

One solution Quadratic Equations 

• In general, the quadratic equation will have two solutions, but there are some special cases where the quadratic equation has only one solution, and some have no solutions. 
• Some quadratic equations have only one solution.  
• For example, 𝒙^𝟐−𝟐𝒙+𝟏=𝟎, 𝒙^𝟐−𝟔𝒙+𝟗=𝟎. 
• Let us see some examples with one solution. 

Example 1: 

Draw the graph for  𝒚 = 𝒙^𝟐−𝟐𝒙+𝟏 𝒂𝒏𝒅 find the solutions for the quadratic equations. 

Firstly, construct a table with x and y values. 

Now, by the table, we have drawn the graph and plotted the points,  

The graph cuts the x-axis(y=0) at x = 1. 

So, from the graph, there is only one solution as the graph touches the x-axis at only one point.  

So, the solution is x = 1. 

Example 2: 

Draw the graph for y = x²−6x+9 and find the solutions for the quadratic equations. 

Firstly, construct a table with x and y values. 

Now, by the table, we have drawn the graph and plotted the points,  

The graph cuts the x-axis(y=0) at x = 3. 

So, from the graph, there is only one solution as the graph touches the x-axis at only one point.  

So, the solution is x = 3. 

No Solution Quadratic Equation 

• Some quadratic equations don’t have a solution, they are also quadratic equations, but they don’t have any real solutions. 
• Let us see an example to understand. 
• Draw the graph for y = 8x²−5x+1 and find the solutions for the quadratic equations. 

 Firstly, construct the table of x and y values. 

Now, by the table, we have drawn the graph and plotted the points,  

The graph doesn’t cut the x-axis(y=0). 

From the graph, there are no solutions as the graph doesn’t touch the x-axis.  

So, there is no solution for the quadratic equation. 

Verifying Solutions 

• How can we verify whether the solution obtained is true or not? 

Yes, we can verify the solution by substituting the roots in the equation. 

Let us understand this from an example. 

The solutions of the quadratic equation x²−7x+12=0  by using graphs are x = 3 and x = 4.  

Let us substitute this values in the equation x²−7x+12=0 … (1) 

Let us take x = 3 

We get, L.H.S = (3)2−7(3)+12, 

L.H.S = 9−21+12,

L.H.S = 0 = R.H.S  

When x = 4 

We get, L.H.S = (4)2−7(4)+12,

L.H.S = 16−28+12,

L.H.S = 0 = R.H.S  

Solution is verified. 

Real-Life Example: 

• A quadratic equation forms a parabola.   
• A football hit forms a path shown in the image. 
• From the graph, we can find how far the football has fallen. 

Discriminant 

• For a quadratic equation ax²+bx+c=0, the expression Δ=b²−4ac is known as the discriminant. 
• Δ is the Greek capital letter delta, which is the symbol used for the discriminant. 
• The solution type of a quadratic equation is determined by a value, which is also known as a discriminant. 
•  If Δ<0, then the quadratic equation has no solution. 
• If Δ=0, then the quadratic equation has only one solution. 
• If Δ>0, then the quadratic equation has two solutions. 

Types of Solutions

Exercise

1. Determine the solutions of the quadratic equation x²-10x+25= 0 by using graphs. Give answers to 1 decimal place where appropriate.
2. Determine the solutions of the quadratic equation x²-3x+5=0 by using graphs. Give answers to 1 decimal place where appropriate.
3. Determine the solutions of the quadratic equation x²-9x+9=0 by using graphs. Give answers to 1 decimal place where appropriate.
4. Determine the solutions of the quadratic equation x²+5x+12= 0 by using graphs. Give answers to 1 decimal place where appropriate.
5. Determine the solutions of the quadratic equation x²-7x+12= 0 by using graphs. Give answers to 1 decimal place where appropriate.
6. Determine the solutions of the quadratic equation 4x²-3x+10= 0 by using graphs. Give answers to 1 decimal place where appropriate.
7. Determine the solutions of the quadratic equation 3x²-2x+1=0 by using graphs. Give answers to 1 decimal place where appropriate.
8. Find the discriminant for the equation x²-7x+12= 0.
9. Find the discriminant for the quadratic equation x²-18x+72 = 0.
10. Determine the type of solutions for the quadratic equation x²+15x+56= 0 by using graphs. Give answers to 1 decimal place where appropriate.

Concept Map

What have we learned

• Quadratic equations using tables
• Quadratic equations using graphs
• Discriminant
• Types of solutions by using discriminant.