As we have seen in the post 'Solution of Quadratic by Graph' we can represent solution of a quadratic equation. Similarly we can represent solution of a cubic equation also by graph. The method is the same we represent a relation between two roots. The relation gives us a graph which has solution as the value of x coordinate and the corresponding y coordinate. In the case of cubic we get an ellipse. Let us first find the relation between two roots of a cubic equation.
Solution of the cubic equation in the above graph is 1,2 and 3. As we can see from the graph that the solution exist where the line x=1 or x=2 or x=3 cuts the curve. When we look at the line x = 1, then we find that the line cuts the curve at y=2 and y=3. Similarly we see that the curve is cut by the line x=2 at y=1 and y=3. So is the case with x=3, where y=1 and y=2. So the solutions of the equation is the points satisfying the condition that the product of the roots is equal to the negative of the constant term divided by coefficient of x3.
Similarly on the guidelines for the previous graph we find that the solution is 1,2 and 4.
When the graph is plotted then the roots of the family of curves ax3 + bx2 + cx + k = 0, where k is parameter is given by the curve which is like the ellipse. If one root is known to be α. Then other two roots can be found by looking at the points where the line x = α intersect the ellipse. By noting down the two values of y where the line x = α intersect we get the three roots. The two values of y and one value of x or two values of x and one value of y are the three roots of the curve.
- Let the equation be ax3 + bx2 + cx + d = 0
- Take two roots of the equation. Let it be x1 and x2.
- Form one equation with one root.
ax13 + bx12 + cx1 + d = 0 ---(i)
Form one more equation with other root.
ax23 + bx22 + cx2 + d = 0 ---(i) - Subtract (ii) from (i) and solve it
(ax13 − ax23) + (bx12 − bx22) + (cx1 − cx2) + d − d = 0
a(x13 − x23) + b(x12 − x22) + c(x1 − x2) = 0
(x1 − x2)[a(x12 + x22 + x1x2) + b(x1 + x2) + c] = 0
Suppose both the roots are different, then
[ax12 + ax22 + ax1x2 + bx1 + bx2 + c] = 0
[ax12 + ax1x2 + bx1 + ax22 + bx2 + c] = 0
[ax12 + x1(ax2 + b) + (ax22 + bx2 + c)] = 0
solving in x1 we get,
x1 = (1/2a)( − (ax2 + b)±√((ax2 + b)2 − 4a(ax22 + bx2 + c)))
x1 = (1/2a)( − (ax2 + b)±√(−3a2x22 − 2abx2 + b2 − 4ac)) You will get a relation in the roots. - Plot the relation on a graph by taking x1 to be y and x2 to be x.
The relation becomes y = (1/2a)( − (ax + b)±√( − 3a2x2 − 2abx + b2 − 4ac))
Divide the relation into two parts one with + sign and other with − sign. Plot them.
Solution of the cubic equation in the above graph is 1,2 and 3. As we can see from the graph that the solution exist where the line x=1 or x=2 or x=3 cuts the curve. When we look at the line x = 1, then we find that the line cuts the curve at y=2 and y=3. Similarly we see that the curve is cut by the line x=2 at y=1 and y=3. So is the case with x=3, where y=1 and y=2. So the solutions of the equation is the points satisfying the condition that the product of the roots is equal to the negative of the constant term divided by coefficient of x3.
When the graph is plotted then the roots of the family of curves ax3 + bx2 + cx + k = 0, where k is parameter is given by the curve which is like the ellipse. If one root is known to be α. Then other two roots can be found by looking at the points where the line x = α intersect the ellipse. By noting down the two values of y where the line x = α intersect we get the three roots. The two values of y and one value of x or two values of x and one value of y are the three roots of the curve.
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