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Ferrari’s method for solving a quartic equation

$$x^4-15x^2-10x+ 24 = 0$$

begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of the form:$$-2bx^2+b^2$$to both sides.

(a) Explain why this is good idea and what it accomplishes.

(b) Use $b= 7$ to find the two quadratic equations that yield the solutions.

$(a)$

I have

$$\begin{align*}x^4-2bx^2+b^2&=15x^2+ 10x-24-2bx^2+b^2\\\\ &=(15-2b)x^2+ 10x+(b^2-24) \end{align*}$$

I then notice that $$x^4-2bx^2+b^2=(x^2-b)^2$$

so we get

$$(x^2-b)^2=(15-2b)x^2+ 10x+(b^2-24)$$

Pluggin in $b=7$ I get

$$(x^2-7)^2=x^2+ 10x+25$$

but I fail to see why this helps with regards to solving for $x$.

Any hints to lean me in the direction would be much appreciated.

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  • 4
    $\begingroup$ $x^2 + 10x + 25 = (x+5)^2$? $\endgroup$ – Anton Grudkin Mar 15 '18 at 2:03
  • $\begingroup$ I don't know how I didn't notice that... $\endgroup$ – Remy Mar 15 '18 at 2:05
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$$(x^2-7)^2=x^2+ 10x+25\implies (x^2-7)^2-(x+5)^2=0$$

$$(x^2-x-12)(x^2+x-2)=0$$

Now solve each quadratic equation for $ x$

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$$(x^2-7)^2 = (x+5)^2$$

$$(x^2-7)^2-(x+5)^2=0$$

$$(x^2-7-x-5)(x^2-7+x+5)=0$$

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  • $\begingroup$ I want for newcomer explain @Mohammad's and Siong's move : Let $A=a_1+a_2$ and $B=b_1+b_2$ $$$$ \begin{align} A^2-B^2=0\\ (A-B)(A+B)=0 \\ (a_1+a_2-(b_1+b_2))(a_1+a_2+b_1+b_2)=0 \\ (a_1+a_2-b_1-b_2)(a_1+a_2+b_1+b_2)=0 \end{align} $\endgroup$ – Krzysztof Myśliwiec Sep 20 '18 at 18:46

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