Question

$\begin{array}{l}\textsf{Se a equa\c{c}\~ao}~\mathsf{~ax^2+bx+c=0,~(a\neq0)~possui~duas~ra\'izes~reais,distintas}\\\mathsf{e~n\~ao~nulas,encontre~a~equa\c{c}\~ao~de~ra\'izes~x_1^2~e~x_2^2.}\end{array}$

Answer 1

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Para não confundir, ao invés de chamar as raízes por $x_1$ e $x_2,$ vou chamá-las por $r_1$ e $r_2.$


Se a equação

$ax^2+bx+c=0\qquad(a\ne 0)$


possui duas raízes reais $r_1$ e $r_2$ distintas e não-nulas, então

$\left\{\! \begin{array}{l} r_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\ r_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a} \end{array} \right.$


ou ainda, as raízes da equação são dadas por

$r=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$


Elevando ambos os lados ao quadrado, obtemos

$r^2=\bigg(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\bigg)^2\\\\\\ r^2=\dfrac{\big(\!\!-b\pm\sqrt{b^2-4ac}\big)^2}{(2a)^2}\\\\\\ r^2=\dfrac{\big(b\mp\sqrt{b^2-4ac}\big)^2}{(2a)^2}\\\\\\ r^2=\dfrac{b^2\mp2b\sqrt{b^2-4ac}+\big(\sqrt{b^2-4ac}\big)^2}{4a^2}$

$r^2=\dfrac{b^2\mp 2b\sqrt{b^2-4ac}+b^2-4ac}{4a^2}\\\\\\ r^2=\dfrac{2b^2-4ac\mp 2b\sqrt{b^2-4ac}}{4a^2}\\\\\\ r^2=\dfrac{\diagup\!\!\!\! 2\cdot \big(b^2-2ac\mp b\sqrt{b^2-4ac}\big)}{\diagup\!\!\!\! 2\cdot 2a^2}\\\\\\ r^2=\dfrac{b^2-2ac\mp b\sqrt{b^2-4ac}}{2a^2}$


Portanto,

$\left\{\! \begin{array}{l} r_1^2=\dfrac{b^2-2ac+b\sqrt{b^2-4ac}}{2a^2}\\\\ r_2^2=\dfrac{b^2-2ac-\sqrt{b^2-4ac}}{2a^2} \end{array} \right.$

__________


A equação quadrática que possui $r_1^2$ e $r_2^2$ como raízes é

$(x-r_1^2)(x-r_2^2)=0\\\\ x^2-r_1^2 x-r_2^2 x+r_1^2r_2^2=0\\\\ x^2-(r_1^2+r_2^2)x+r_1^2r_2^2=0$


Para facilitar a manipulação dos termos, multiplica os dois lados da equação por $4a^4,$ pois $a\ne 0:$

$4a^4\cdot \big[x^2-(r_1^2+r_2^2)x+r_1^2r_2^2\big]=4a^4\cdot 0\\\\ 4a^4x^2-4a^4\cdot(r_1^2+r_2^2)x+4a^4\cdot r_1^2r_2^2=0\qquad\quad\mathbf{(i)}$


Porém, temos que

$\bullet~~4a^4\cdot (r_1^2+r_2^2)\\\\ =4a^4\cdot \bigg(\dfrac{b^2-2ac+b\sqrt{b^2-4ac}}{2a^2}+\dfrac{b^2-2ac-b\sqrt{b^2-4ac}}{2a^2}\bigg)\\\\\\ =4a^4\cdot \dfrac{b^2-2ac+b\sqrt{b^2-4ac}+b^2-2ac-b\sqrt{b^2-4ac}}{2a^2}\\\\\\ =4a^4\cdot \dfrac{\diagup\!\!\!\! 2(b^2-2ac)}{\diagup\!\!\!\! 2a^2}\\\\\\ \therefore~~4a^4\cdot (r_1^2+r_2^2)=4a^2\cdot (b^2-2ac)\qquad\quad\mathbf{(ii)}$


$\bullet~~4a^4\cdot r_1^2r_2^2\\\\ =4a^4\cdot \dfrac{b^2-2ac+b\sqrt{b^2-4ac}}{2a^2}\cdot \dfrac{b^2-2ac-b\sqrt{b^2-4ac}}{2a^2}\\\\\\ =4a^4\cdot \dfrac{\big(b^2-2ac+b\sqrt{b^2-4ac}\big)\cdot \big(b^2-2ac-b\sqrt{b^2-4ac}\big)}{2a^2\cdot 2a^2}\\\\\\ =4a^4\cdot \dfrac{(b^2-2ac)^2-\big(b\sqrt{b^2-4ac}\big)^2}{4a^4}$

$=(b^2-2ac)^2-\big(b\sqrt{b^2-4ac}\big)^2\\\\ =b^4-4ab^2c+4a^2c^2-b^2\cdot (b^2-4ac)\\\\ =\diagup\!\!\!\! b^4-~~\diagdown\!\!\!\!\!\!\!\!\! 4ab^2c+4a^2c^2-\diagup\!\!\!\! b^4+~~\diagdown\!\!\!\!\!\!\!\!\! 4ab^2c\\\\\\ \therefore~~4a^4r_1^2r_2^2=4a^2c^2\qquad\quad\mathbf{(iii)}$


Substituindo em $\mathbf{(i)},$ ficamos com

$4a^4x^2-\underbrace{4a^4\cdot(r_1^2+r_2^2)}x+\underbrace{4a^4\cdot r_1^2r_2^2}=0\\\\\\ 4a^4x^2-4a^2\cdot (b^2-2ac)x+4a^2c^2=0\\\\ 4a^2\cdot a^2x^2-4a^2\cdot (b^2-2ac)x+4a^2c^2=0\\\\ 4a^2\cdot \big[a^2x^2-(b^2-2ac)x+c^2\big]=0\\\\\\ \therefore~~\boxed{\begin{array}{c} a^2x^2-(b^2-2ac)x+c^2=0 \end{array}}\quad\longleftarrow\quad\textrm{esta \'e a equa\c{c}\~ao procurada}\\\\\\ \textrm{pois }a\ne 0.$


Dúvidas? Comente.


Bons estudos! :-)