Question

1_ Determine as soluções, caso existam, para as seguintes equações. (a) 3x² – 15x + 12 = 0 (b) 2x² – 14x + 20 = 0 (c) 4x² + 4x – 24 = 0 (d) x² + 3x – 18 = 0 (e) x² - 10x + 40 = 0


por favor gente me ajudem! ;(​

Answer 1

Sua questão:

Determine as soluções, caso existam, para as seguintes equações.

(a) 3x² – 15x + 12 = 0

(b) 2x² – 14x + 20 = 0

(c) 4x² + 4x – 24 = 0

(d) x² + 3x – 18 = 0

(e) x² - 10x + 40 = 0

Resolução:

( A )

$\large\begin{array}{lr} \sf 3x^2 - 15x + 12 = 0 \left\{\begin{array}{ll}\sf a=3\\\sf b=-15\\\sf c=12\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = (-15)^2 -4*3*12\\\\\sf \Delta = 225-144\\\\\sf \underline{\boxed{\red{\sf \Delta =81}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-(-15) \pm \sqrt{81} }{2*3} \\\\ \sf x= \dfrac{ 15 \pm 9 }{6} \\\\ \sf x'= \dfrac{24}{6} = \underline{\boxed{\red{\sf x' = 4}}} \\\\ \sf x''= \dfrac{6}{6} = \underline{\boxed{\red{\sf x'' = 1}}} \end{array}$

___________#____________

( B )

$\large\begin{array}{lr} \sf 2x^2 - 14x + 20 = 0\left\{\begin{array}{ll}\sf a=2\\\sf b=-14\\\sf c=20\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = (-14)^2 -4*2*20\\\\\sf \Delta = 196-160\\\\\sf \underline{\boxed{\red{\sf \Delta =36}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-(-14) \pm \sqrt{36} }{2*2} \\\\ \sf x= \dfrac{ 14 \pm 6 }{4} \\\\ \sf x'= \dfrac{20}{4} = \underline{\boxed{\red{\sf x' = 5}}} \\\\ \sf x''= \dfrac{8}{4} = \underline{\boxed{\red{\sf x'' = 2}}} \end{array}$

___________#____________

( C )

$\large\begin{array}{lr} \sf 4x^2 + 4x - 24 = 0\left\{\begin{array}{ll}\sf a=4\\\sf b=4\\\sf c=-24\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = 4^2 -4*4*(-24)\\\\\sf \Delta = 16+384\\\\\sf \underline{\boxed{\red{\sf \Delta =400}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-4 \pm \sqrt{400} }{2*4} \\\\ \sf x= \dfrac{ -4 \pm 20 }{8} \\\\ \sf x'= \dfrac{16}{8} = \underline{\boxed{\red{\sf x' = 2}}} \\\\ \sf x''= \dfrac{-24}{8} = \underline{\boxed{\red{\sf x'' = -3}}} \end{array}$

___________#____________

( D )

$\large\begin{array}{lr} \sf x^2+ 3x - 18 = 0\left\{\begin{array}{ll}\sf a=1\\\sf b=3\\\sf c=-18\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = 3^2 -4*1*(-18)\\\\\sf \Delta = 9+72\\\\\sf \underline{\boxed{\red{\sf \Delta =81}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-3\pm \sqrt{81} }{2*1} \\\\ \sf x= \dfrac{ -3 \pm 9 }{2} \\\\ \sf x'= \dfrac{6}{2} = \underline{\boxed{\red{\sf x' = 3}}} \\\\ \sf x''= \dfrac{-12}{2} = \underline{\boxed{\red{\sf x'' = -6}}} \end{array}$

___________#____________

( E )

$\large\begin{array}{lr} \sf x^2 - 10x + 40 = 0\left\{\begin{array}{ll}\sf a=1\\\sf b=-10\\\sf c=40\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = 10^2 -4*1*(40)\\\\\sf \Delta = 100-160\\\\\sf \underline{\boxed{\red{\sf \Delta =-60}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-(-10)\pm 2\sqrt{-15} }{2} \\\\ \sf x= 5\pm \sqrt{-15} \\\\ \sf x'=\underline{\boxed{\red{\sf 5 + (\sqrt{15})i}}} \\\\ \sf x''= \underline{\boxed{\red{\sf 5 - (\sqrt{15})i}}} \end{array}$

Espero ter ajudado.

Bons estudos.

• Att. FireClassis.