Question

$\frac{5 x^{2}+20x}{4x- x^{2} } =$

x²+2√2x-30=0

Answer 1

[tex] x^{2} +2 \sqrt{2}x-30=0 \\ \Delta= (2 \sqrt{2} )^{2}-4*1*(-30)=8+120=128 \\ x_{1}= \frac{-b- \sqrt{\Delta}}{2a}= \frac{-2 \sqrt{2} - \sqrt{128} }{2}=\frac{-2 \sqrt{2} - 8\sqrt{2} }{2}= \frac{-10 \sqrt{2} }{2} =-5 \sqrt{2} \\ x_{2} =\frac{-b+ \sqrt{\Delta}}{2a}= \frac{-2 \sqrt{2} + \sqrt{128} }{2}=\frac{-2 \sqrt{2} + 8\sqrt{2} }{2}= \frac{6 \sqrt{2} }{2}=3 \sqrt{2} [/tex]
S={-5 radical din 2; 3radical din 2}
$\frac{ 5x^{2} +20x}{4x- x^{2} } = \frac{5x(x+4)}{x(4-x)}= \frac{5(x+4)}{4-x}$