Question

Encontre as raízes da equação e coloque no conjunto solução:
4x² + 20x=0
x² + 3x=0
4x² + 5x=0
50 x²+ 100x=0
x² -81=0
2x² - 200=0
x² - 144 =0
3 x² - 48 = 0

Answer 1

$\boxed{\begin{array}{lr} a)\\4x^2+20x=0 \end{array}}$

• Primeiro temos que dividir os dois lados da equação pelo mesmo termo.

$\boxed{\begin{array}{lr} 4x^2+20x=0\\4(x^2+5x)=0\\x^2+5x=0\\ \end{array}}$

• Agora temos que ter os coeficientes, que são eles...

A = 1

B = 5

C = 0

• Agora resolvendo com a formula de bhaskara.

$\boxed{\begin{array}{lr} x=\dfrac{-b\pm\sqrt{\Delta}}{2.a} \end{array}} \to \boxed{\begin{array}{lr} \Delta= b^2-4.a.c \end{array}}$

• Trocando as letras pelos valores de seus coeficientes.
• E achando o valor de Delta.

$\boxed{\begin{array}{lr} x=\dfrac{-5\pm\sqrt{5^2-4.1.0}}{2.1} \end{array}}\\\\\sf \bf Resolvendo\\\\\boxed{\begin{array}{lr} x=\dfrac{-5\pm\sqrt{25+0}}{2} \end{array}}\\\\\\\boxed{\begin{array}{lr}\boxed{\begin{array}{lr} x=\dfrac{-5\pm5}{2} \end{array}} \end{array}}$

• Agora é preciso retirar o mais ou menos.

$\boxed{\begin{array}{lr} x'=\dfrac{-5+5}{2} \end{array}}\to \boxed{\begin{array}{lr} x'=\dfrac{0}{2} \end{array}}\to \boxed{\begin{array}{lr} x'=0 \end{array}}\\\\\boxed{\begin{array}{lr} x''=\dfrac{-5-5}{2} \end{array}}\to \boxed{\begin{array}{lr} x''=\dfrac{-10}{2} \end{array}} \to \boxed{\begin{array}{lr} x''=-5 \end{array}}$

Resposta;

$S=\{0,-5\}$

=============

$\boxed{\begin{array}{lr} b)\\x^2+3x=0 \end{array}}$

$A=1\\B=3\\C=0$

$\boxed{\begin{array}{lr} x=\dfrac{-3\pm\sqrt{3^2-4.1.0}}{2} \end{array}}\to \boxed{\begin{array}{lr} x=\dfrac{-3\pm \backslash\!\!\!\!\!\sqrt{3\backslash\!\!\!^2\backslash\!\!\!\!+ \backslash\!\!\!0}}{2} \end{array}} \to \boxed{\begin{array}{lr} x=\dfrac{-3\pm3}{2} \end{array}}\\\\\\\boxed{\begin{array}{lr} x'=\dfrac{-3+3}{2} \end{array}}\to \boxed{\begin{array}{lr} x'=\dfrac{0}{2} \end{array}} \to \boxed{\begin{array}{lr} x'=0 \end{array}}$

$\boxed{\begin{array}{lr} x''=\dfrac{-3-3}{2} \end{array}}\to \boxed{\begin{array}{lr} x''=\dfrac{-6}{2}\end{array}}\to\boxed{\begin{array}{lr} x''=-3 \end{array}}$

Resposta;

$S=\{0,-3\}$

==============

$\boxed{\begin{array}{lr} c)\\4x^2+5x=0 \end{array}}\\\boxed{\begin{array}{lr} A=4\\B=5\\C=0 \end{array}}\\\boxed{\begin{array}{lr} x=\dfrac{-5\pm\sqrt{5^2-4.4.0}}{2.4} \end{array}}\to\boxed{\begin{array}{lr} x=\dfrac{-5\pm\backslash\!\!\!\!\!\sqrt{5\backslash\!\!\!^2}}{8} \end{array}}\to\boxed{\begin{array}{lr} x=\dfrac{-5\pm5}{8} \end{array}}\\\\\\\boxed{\begin{array}{lr} x'=\dfrac{-5+5}{8} \end{array}}\to \boxed{\begin{array}{lr} x'=\dfrac{0}{8} \end{array}}\to \boxed{x'=0}$

$\boxed{\begin{array}{lr} x''=\dfrac{-5-5}{8} \end{array}}\to \boxed{\begin{array}{lr} x''=\dfrac{-10}{8} \end{array}}\to \boxed{\begin{array}{lr} x''=\dfrac{-5}{4} \end{array}}$

Resposta;

$S=\{0,-5/4\}$

===================

$\boxed{\begin{array}{lr} d)\\50x^2+100x=0 \end{array}}\\\boxed{\begin{array}{lr} 50(x^2+2x)=0\\x^2+2x=0 \end{array}}$

$\boxed{\begin{array}{lr} A=1\\B=2\\C=0 \end{array}}$

$\boxed{\begin{array}{lr} x=\dfrac{-2\pm\sqrt{2^2-4.1.0}}{2.1} \end{array}}\to\boxed{\begin{array}{lr} x=\dfrac{-2\pm\sqrt{2^2}}{2} \end{array}}\to\boxed{\begin{array}{lr} x=\dfrac{-2\pm2}{2} \end{array}}\\\\\\\boxed{\begin{array}{lr} x'=\dfrac{-2+2}{2} \end{array}}\to\boxed{\begin{array}{lr} x'=\dfrac{0}{2} \end{array}}\to\boxed{\begin{array}{lr} x'=0 \end{array}}\\\boxed{\begin{array}{lr} x''=\dfrac{-2-2}{2} \end{array}}\to\boxed{\begin{array}{lr} x''=\dfrac{-4}{2} \end{array}}\to\boxed{x''=-2}$

$S=\{0,-2\}$

==============

$\boxed{\begin{array}{lr} e)\\x^2-81=0 \end{array}}\\\boxed{\begin{array}{lr} x^2-81=0\\x^2=81\\x=\sqrt{81}\\x=\pm \ 9\\\boxed{\begin{array}{lr} x'=+\ 9\\x''=- \ 9 \end{array}} \end{array}}$

Resposta;

S = {9,-9}

===============

$\boxed{\begin{array}{lr} f)\\2x^2-200=0 \end{array}}\\\boxed{\begin{array}{lr} 2x^2-200=0\\2x^2=200\\2x=\sqrt{200}\\x=\sqrt{\dfrac{200}{2}}\\x=\sqrt{100}\\x=10\\x=\pm\ 10\\\boxed{\begin{array}{lr} x'=+\ 10\\x''=-\ 10 \end{array}} \end{array}}$

S = {10,-10}

===============

$\boxed{\begin{array}{lr} g)\\x^2-144=0 \end{array}}\\\boxed{\begin{array}{lr} x^2-144=0\\x^2=144\\x=\sqrt{144}\\x=\pm12\\\boxed{\begin{array}{lr} x'=12\\x''=-12 \end{array}} \end{array}}$

S = {12,-12}

================

$\boxed{\begin{array}{lr} h)\\3x^2-48=0 \end{array}}\\\boxed{\begin{array}{lr} 3x^2-48=0\\3x^2=48\\3x=\sqrt{48}\\x=\sqrt{\dfrac{48}{3}}\\x=\sqrt{16}\\x=\pm4\\\boxed{\begin{array}{lr} x'=4\\x''=-4 \end{array}} \end{array}}$

S = {4,-4}

===========

Saiba Mais em;

brainly.com.br/tarefa/46139021

$|\underline{\overline{\mathcal{\boldsymbol{\LaTeX}}}}|\\|\underline{\overline{\mathcal{\boldsymbol{\mathbbe\mathcal{{ATT:JL\ \ \heartsuit}}}}}}|$