Question

Cine ma poate ajuta???? VA ROOOG

Answer 1

[tex] a) \sqrt{6-4x- x^{2} }=x+4|^2\\ 6-4x- x^{2} =(x+4)^2\\ 6-4x-x^2= x^{2} +8x+16\\ 6-4x- x^{2} - x^{2} -8x-16=0\\-2x^{2} -12x-10=0|:(-2)\\ x^{2} +6x+5=0\\ \Delta=b^2-4ac=6^2-4*1*5=36-20=16\\x_1= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-6- \sqrt{16} }{2*1}= \frac{-6-4}{2}= \frac{-10}{2}=-5\\ x_2= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-6+ \sqrt{16} }{2*1}= \frac{-6+4}{2}= \frac{-2}{2}=-1\\ [/tex]
[tex]b) 2 \sqrt{x+5}=x+2|:2\\ \sqrt{x+5}= \frac{x+2}{2}|^2\\ x+5= \frac{(x+2)^2}{2^2} \\ 2^2*(x+5)=(x+2)^2\\ 4(x+5)= x^{2} +2*x*2+2^2\\4x+20 =x^{2} +4x+4\\ 4x+20- x^{2} -4x-4=0\\ - x^{2} +16=0|*(-1)\\ x^{2} -16=0\\ x^{2} -4^2=0\\(x-4)(x+4)=0\\ x-4=0\\sau\\ x+4=0\\ x_1=4\\x_2=-4\\[/tex]
[tex]c) \sqrt{4-x}+ \sqrt{5+x}=3\\ \sqrt{5+x} =3- \sqrt{4-x}|^2\\ 5+x=(3- \sqrt{4-x} )^2\\ 5+x=3^2-2*3* \sqrt{4-x}+( \sqrt{4-x} )^2\\ 5+x=9-6 \sqrt{4-x}+4-x\\ 5+x=13-x-6 \sqrt{4-x}\\5+x-13+x=-6 \sqrt{4-x}\\2x-8=-6 \sqrt{4-x}|*(-1)\\8-2x=6 \sqrt{4-x}|:6\\ \frac{8-2x}{6}= \frac{6\sqrt{4-x}}{6} \\ \frac{2(4-x)}{6}= \sqrt{4-x}\\ \frac{4-x}{3}= \sqrt{4-x}|^2\\ \frac{(4-x)^2}{3^2}=( \sqrt{4-x})*2\\ \frac{ (4-x)^2}{9}=4-x\\ (4-x)^2=9(4-x)\\ 4^2-2*4*x+ x^{2} =36-9x\\ 16-8x+ x^{2} =36-9x\\ [/tex]
[tex] 16-8x+ x^{2} -36+9x=0\\ x^{2} +x-20=0\\\Delta=b^2-4ac=1^2-4*1*(-20)=1+80=81\\ x_1= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-1- \sqrt{81} }{2*1}= \frac{-1-9}{2} = \frac{-10}{2}=-5\\ x_2= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-1+ \sqrt{81} }{2*1}= \frac{-1+9}{2} = \frac{8}{2}=4\\ [/tex]