Question

3
.1. Solve the equation : x°-15-126=0 by Cardan's Method.​

Answer 1

Answer:

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Answer 2

To Solve : For Values of x such that

$f(x) = {x}^{3} - 15x - 126 = 0$

Here,

$Degree [f(x),x] = 3 \\ \: \: \: \: \: \: \: \: \: Coefficient[f(x),x] = 0$

Therefore, x must be a binomial.

Calculating the nature of roots:

$\\ Discriminant[f(x) , x] = - 27( - 126)^{2} - 4( { - 15})^{3}$

$\triangle = - 415152 < 0$

$\implies x _{1} \in ℝ$

Now,

Let x being a binomial.

$: = \: x = u + v$

$(u + v) ^{3} - 15(u + v) - 126 = 0$

$\\ {u}^{3} + {v}^{3} +3uv(u + v) - 15(u + v) - 126 = 0$

$\\ ( {u}^{3} + {v}^{3} - 126) + (u + v)(3uv - 15) = 0$

Therefore, Simutaneously, we must conclude that

$\implies {u}^{3} - 126 + {v}^{3} = 0$

$\implies 3uv - 15 = 0$

$⇔uv = 5$

Now, Multiplying First Equation by (u)³ on both sides and substituting using second equation

$( {u}^{6} ) - 126( {u}^{3} ) + 125 = 0$

${u}^{3} = \frac{126 \pm \sqrt{(126)^{2} - 4 (125)} }{2}$

${u}^{3} = 63 \pm \frac{ \sqrt{ {124}^{2} } }{2}$

${u}^{3} = 63 \pm62$

But, As u and v are indistinguishable. And, u≠v

Therefore,

$\: \: \: \: {u}^{3} = 63 + 62 = 125 \: \\ {v}^{3} = 63 - 62 = 1 \:$

And, Thus,

$u = 5 \: , \: 5 \omega \: , \: 5 {\omega}^{2}$

$v = 1 \: , \: \omega ^{2} \: , \: \omega$

Therefore, All the values of x are:

$\implies x_{1} = 6$

$\implies x_{2} = - 3 + 2 \sqrt{3} i$

$\implies x_{3} = - 3 - 2\sqrt{3} i$