Question

Por meio da fórmula de bhaskara, determine as raízes de cada equação:

a) x²-6x+5=0

b) 3x²+4x+1=0

c) x²+3x-28=0

d) -x²+9x-20=0

e) 3x²-4x+2=0​​

Answer 1

Explicação passo-a-passo:

$a) {x}^{2} - 6x + 5 = 0 \\ delta = {( - 6)}^{2} - 4 \times 1 \times 5 \\ delta = 6 - 20 \\ delta = 16 \\ \frac{ - {( - 6)}+ - \sqrt{16} }{2 \times 1} \\ \frac{6 + - 4}{2} \\ {x}^{1} = \frac{6 + 4}{2} = \frac{10}{2} = 5 \\ {x}^{2} = \frac{6 - 4}{2} = \frac{2}{2} = 1$

$b)3 {x}^{2} + 4x + 1 = 0 \\ delta = {4}^{2} - 4 \times 3 \times 1 \\ delta = 16 - 12 \\ delta = 4 \\ \frac{ - 4 + - \sqrt{4} }{2 \times 3} \\ \frac{ - 4 + - 2}{6} \\ {x}^{1} = \frac{ - 4 + 2}{6} = \frac{ - 2}{6} = - 0.333 \\ {x}^{2} = \frac{ - 4 - 2}{6} = \frac{ - 6}{6} = - 1$

tá aí irmão o resto, bons estudos!

Answer 2

$\Large\boxed{\begin{array}{l}\tt a)~\sf x^2-6x+5=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-6)^2-4\cdot1\cdot5\\\sf\Delta=36-20\\\sf\Delta=16\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-6)\pm\sqrt{16}}{2\cdot1}\\\\\sf x=\dfrac{6\pm4}{2}\begin{cases}\sf x_1=\dfrac{6+4}{2}=\dfrac{10}{2}=5\\\\\sf x_2=\dfrac{6-4}{2}=\dfrac{2}{2}=1\end{cases}\\\sf S=\{1,5\}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf 3x^2+4x+1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=4^2-4\cdot3\cdot1\\\sf\Delta=16-12\\\sf\Delta=4\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-4\pm\sqrt{4}}{2\cdot3}\\\\\sf x=\dfrac{-4\pm2}{6}\begin{cases}\sf x_1=\dfrac{-4+2}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\\\\\sf x_2=\dfrac{-4-2}{6}=-\dfrac{6}{6}=-1\end{cases}\\\\\sf S=\bigg\{-\dfrac{1}{3},-1\bigg\}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)~\sf x^2+3x-28=0\\\sf\Delta=b^2-4ac\\\sf\Delta=3^2-4\cdot1\cdot(-28)\\\sf\Delta=9+112\\\sf\Delta=121\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-3\pm\sqrt{121}}{2\cdot1}\\\\\sf x=\dfrac{-3\pm11}{2}\begin{cases}\sf x_1=\dfrac{-3+11}{2}=\dfrac{8}{2}=4\\\\\sf x_2=\dfrac{-3-11}{2}=-\dfrac{14}{2}=-7\end{cases}\\\sf S=\{-7,4\}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt d)~\sf -x^2+9x-20=0\cdot(-1)\\\sf x^2-9x+20=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-9)^2-4\cdot1\cdot20\\\sf\Delta=81-80\\\sf\Delta=1\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-9)\pm\sqrt{1}}{2\cdot1}\\\\\sf x=\dfrac{9\pm1}{2}\begin{cases}\sf x_1=\dfrac{9+1}{2}=\dfrac{10}{2}=5\\\\\sf x_2=\dfrac{9-1}{2}=\dfrac{8}{2}=4\end{cases}\\\sf S=\{4,5\}\end{array}}$$\Large\boxed{\begin{array}{l}\tt e)~\sf 3x^2-4x+2=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot3\cdot2\\\sf\Delta=16-24\\\sf\Delta=-8<0\\\sf N\tilde ao~existem~ra\acute izes~reais.\end{array}}$