Question

Solve the equation : x3 -15-126=0 by Cardan’s Method.

Answer 1

Answer:

64x³–48x²+12x–1

Cubic in normal form: x³–0.75x²+0.1875x–0.015625

x³+ax²+bx+c [1]

Substitute x=t–a/3, to eliminate the x² term

New equation t³+pt+q=0

Where p=b–a²/3, and q=c+ (2a³–9ab)/27

p=0 q=0

So the new equation is:

t3+0t +0 [2]

As p and q are zero, then all the roots are real and equal to –a/3 (0.25)

In summary, the roots for

Cubic: 64x³–48x²+12x–1 are:

x1=0.25 (Remainder=0)

x2=0.25 (Remainder=0)

x3=0.25 (Remainder=0)

The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+6x²+11x+6

Cubic in normal form: x³+6x²+11x+6

x³+ax²+bx+c [1]

Substitute x=t–a/3, to eliminate the x² term

New equation t³+pt+q=0

Where p=b–a²/3, and q=c+ (2a³–9ab)/27

p=–1 q=0

So the new equation is:

t3–1t +0 [2]

x1=–a/3, x2=–a/3± √(–p)

As p<0, then all the roots are real

In summary, the roots for

x³+6x²+11x+6 are:

x1=–2 (Remainder=0)

x2=–1 (Remainder=0)

x3=–3 (Remainder=0)