Question

Formula de Baskara Fracionaria​

Answer 1

d)

$\frac{ {x}^{2} }{4} - \frac{x}{3} + \frac{1}{9} = 0 \times (36) \\ 9 {x}^{2} - 12x + 4 = 0$

${(3x - 2)}^{2} = 0 \\ 3x - 2 = 0 \\ 3x = 2 \\ x = \frac{2}{3}$

e)

${x}^{2} - 3 = \frac{x - 3}{6} \times( 6) \\ 6 {x}^{2} - 18 = x - 3 \\ 6 {x}^{2} - x - 18 + 3 = 0 \\ 6 {x}^{2} - x - 15 = 0$

6x²+9x-10x-15=0

3x(2x+3)-5(2x+3)=0

(2x+3)(3x-5)=0

$2x + 3 = 0 \\ 2x = - 3 \\ x = - \frac{3}{2}$

$3x - 5 = 0 \\ 3x = 5 \\ x = \frac{5}{3}$

f)

$\frac{ {x}^{2} - 5x}{6} + 1 = \frac{2x + 11}{3} \times (6) \\ {x}^{2} - 5x + 6 = 4x + 22 \\ {x }^{2} - 5x - 4x + 6 -22 = 0$

${x}^{2} - 9x - 16 = 0$

∆=81+64=145

$x = \frac{9± \sqrt{145} }{2} \\ x' = \frac{9 + \sqrt{145} }{2} \\ x''</p><p> = \frac{9 - \sqrt{145} }{2}$