Solve cubic equations by drawing appropriate lines on graphs
Solving cubic equations by drawing appropriate lines on graphs is similar to solving quadratic equations by using graphs.
Example: Solve the cubic equation x^{3} + 5x^{2} +2x 10, using a graph.
First off, we need to draw the graph. We do this by setting our equation to y. y = x^{3} + 5x^{2} +2x 10
Now we can use a table of values to get the values of y we'll need to plot the graph.
To get our table of values, we're going find what is y is for values from x = 5 to x=5. Since we have a cubic graph, we're expecting our y values to off low, increases, then decrease then increase again. (remember the wiggly shape of a cubic graph!)
x  5  4  3  2  1  0  1  2  3  4  5 
y  20  2  2  2  8  10  2  22  68  142  250 
The table above shows the values obtained for the cubic equationy y = x^{3} + 5x^{2} +2x 10. Now if you making this table during an exam, and saw that the values started shooting up after x=3. You would stop there. There is no point trying to plot (5, 250) as your scale on the y axis will make the overall graph look tiny!
This is what our cubic should look like. Remember to make the lines between each point curvy and not straight!
To solve our cubic equation, we need to see where our graph cuts the x axis (y =0 )
From the graph we can see that the graph cuts the x axis at 3 points. At x = 3.75, x= 2.35 and at x= 1.1
Remember to check your results, to see if 1) Your graph is correct and 2) if the points you've selected are the correct ones.
Feeding our values back into x^{3} + 5x^{2} +2x 10 should give us a value near to 0 (since our values are estimations and not 100% accurate)
For x = 3.75:
3.75^{3} +(5 x (3.75)^{2}) + 2 x 3.75 10 = 0.08 (2 d.p.)
For x = 2.35:
2.35^{3} +(5 x (2.35)^{2}) + 2 x 2.35 10 = 0.07 (2 d.p.)
For x = 1.1
1.1^{3} +(5 x (1.1)^{2}) + 2 x 1.1 10 = 0.42 (2 d.p.)
All our estimates are close to 0, so we can say the solutions to the equation are x= 3.75 or x=2.35 or x =1.1
Example 2: Solve x^{3}+6x^{2}+5x6 = x+2, using a graph
For this second example, our equation is equal to a linear equation instead of 0. This time we'll have to draw two graphs y = x^{3}+6x^{2}+5x6 and y=x+2. Then by looking at the graph and seeing where the two intercept, we'll be able to tell what ours solutions should be.
We need to find the values of y for our cubic equation. Again we'll take the values between x = 5 and 5
x  5  4  3  2  1  0  1 
y  6  6  6  0  6  6  6 
x got very large after x=1 so the rest was omitted.
Since y = x+2 is a line, we'll only need two points. We know we need the line to cover the space between x=5 and x=1
x  5  1 
y  3  3 
The estimates of where the graphs intersect each other are at: x = 4.8, x= 2 and x =0.8.
Again, we need to check these results to make sure we've done everything correctly.
For x = 4.8
(4.8)^{3}+(6 x (4.8)^{2}) + 5 x (4.8)  6 = 4.8+2
=> 2.35 = 2.8
For x = 2
2^{3} + (6 x 2^{2}) + 5 x (2)  6 = 2 +2
0 = 0
For x = 0.8
(0.8)^{3}+(6 x (0.8)^{2}) + 5 x (0.8)  6 = 0.8+2
2.35 = 2.8
Our values are fairly close to one another, so we can say that the solutions for x^{3}+6x^{2}+5x6 = x+2 are x= 4.8 or x =2 or x = 0.8
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