Question

resolva as equações do 2° grau usando a fórmula de Bhaskara

C) x²-2x+1=0

D) x²+3x-10=0​

Answer 1

$\large\boxed{\begin{array}{l} \sf{quest\tilde{a}o \:C : } \\ \\ \rm \: x {}^{2} - 2x + 1 = 0 \\ \\ \rightarrow \begin{cases} \rm \: a = 1 \\ \rm \: b = - 2 \\ \rm \: c = 1\end{cases} \\ \\ \rm \Delta = b {}^{2} - 4ac \\ \Delta = ( - 2) {}^{2} - 4 \: . \: 1 \: . \: 1 \\ \Delta = 4 - 4 \\\Delta = 0 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ \rm \: x = \dfrac{ - ( - 2)}{2 \: . \: 1} \\ \\ \rm \: x = \dfrac{2}{2} \\ \\ \boxed{ \rm{x = 1}}\end{array}}$

$\large\boxed{\begin{array}{l} \sf{quest\tilde{a}o \: D : } \\ \\ \rm \: x {}^{2} + 3x - 10 = 0 \\ \\ \rightarrow \begin{cases} \rm \: a = 1 \\ \rm \: b = 3 \\ \rm \: c = - 10 \end{cases} \\ \\ \rm \Delta = b {}^{2} - 4ac \\ \Delta = 3 {}^{2} - 4 \: . \: 1 \: . \: ( - 10) \\ \Delta = 9 + 40 \\\Delta = 49 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ \rm \: x = \dfrac{ - 3 \pm \sqrt{49} }{2 \: . \: 1} \\ \\ \rm \: x = \dfrac{ - 3 \pm7}{2} \begin{cases} \rm \: x_1 = \dfrac{ - 3 + 7}{2} = \dfrac{4}{2} = \boxed{2} \\ \\ \rm \: x_2 = \dfrac{ - 3 - 7}{2} = \dfrac{ - 10}{2} = \boxed{ - 5} \end{cases}\end{array}}$

Answer 2

Explicação passo-a-passo:

x²-2x+1=0

a=1

b=-2

c=1

∆=b²-4ac

∆=(-2)²-4*1*1

∆=4-4

∆=0

-b±√∆/2a

2±√0/2*1

2±0/2

x¹=2+0/2=2/2=>1

x²=2-0/2=2/2=>1

x²+3x-10=0

a=1

b=3

c=-10

∆=b²-4ac

∆=3²-4*1*-10

∆=9+40

∆=49

-b±√∆/2a

-3±√49/2*1

-3±7/2

x¹=-3+7/2=4/2=>2

x²=-3-7/2=-10/2=>-5