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What is the difference between a linear equation and a cubic equation?

By Filsan on the 10th of January, 2013

3 Answers

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    The difference between a linear equation and a cubic equation:

    Each equation gets its form based on the highest degree or exponent, of the variable. For instance, in the case where y = x³ -- 6x + 2, the degree of 3 gives this equation the name "cubic." Any equation that has a degree no higher than 1 receives the name "linear." Otherwise, we call an equation "nonlinear," whether it is quadratic, a sine-curve or in any other form.

     



    Refine By Bridget Gaskin on the 17th of January, 2013

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    Each equation gets its form based on the highest degree, or exponent, of the variable.

    For instance,

    1. y = x³ -- 6x + 2, the degree(Highest power of x) of 3 gives this equation the name "cubic."

    2. Any equation that has a degree no higher than 1 such as x+1=y receives the name "linear."

    Refine By Cecil on the 16th of January, 2013

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    A linear equation is one where the solution set can be drawn as a line. For example, something of the form y = mx + b. Since m and b are constants, the set of point (x,y) that are solutions to the equation forms a line. The important difference in the equation is that both x and y are not raised to any powers.

    A quadratic equation is one where the solution set can be drawn as a parabola which is a special curve shape. The equation generally involves either x or y raised to the second power. For example, something of the form y = ax² + bx + c or x = ay² + by + c where a, b, and c are constants.

     

    Refine By AMRITA CHAKRAVARTY on the 21st of January, 2013

Suggested reading…

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 x3 + 5x2 +2x -10, using a graph. 

First off, we need to draw the graph. We do this by setting our equation to y. y = x3 + 5x2 +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 = x3 + 5x2 +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 x3 + 5x2 +2x -10 should give us a value near to 0 (since our values are estimations and not 100% accurate)

For x = -3.75:

-3.753 +(5 x (-3.75)2) + 2 x -3.75 -10 = 0.08 (2 d.p.)

For x = -2.35:

-2.353 +(5 x (-2.35)2) + 2 x -2.35 -10 = -0.07 (2 d.p.)

For x = 1.1

1.13 +(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  x3+6x2+5x-6 = 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 = x3+6x2+5x-6 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

-23 + (6 x -22) + 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 x3+6x2+5x-6 = x+2 are x= -4.8 or x =-2 or x = 0.8

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