Question

Problema 6 Multumesc anticipat

Answer 1

a)

$x^2+10x+24=0$

$\frac{-10+\sqrt{10^2-4\cdot \:1\cdot \:24}}{2\cdot \:1} =$

$\sqrt{10^2-4\cdot \:1\cdot \:24}=\sqrt{4}$

$=\frac{-10+\sqrt{4}}{2\cdot \:1} =\frac{-10+\sqrt{4}}{2} =\frac{-10+2}{2} =\frac{-8}{2} =-\frac{8}{2} =-4$

$\frac{-10-\sqrt{10^2-4\cdot \:1\cdot \:24}}{2\cdot \:1} =\frac{-10-\sqrt{4}}{2\cdot \:1} =\frac{-10-\sqrt{4}}{2} =\frac{-10-2}{2} =\frac{-12}{2} =-\frac{12}{2} =-6$

$x=-4,\:x=-6$

b)

$\left(x+2\right)^2=7-4\sqrt{3}$

$x+2=\sqrt{7-4\sqrt{3}}$

$\sqrt{7-4\sqrt{3}} =\sqrt{3-4\sqrt{3}+4} =\sqrt{1\cdot \:3-4\sqrt{3}+4}=\sqrt{\left(\sqrt{1}\right)^2\left(\sqrt{3}\right)^2-4\sqrt{3}+\left(\sqrt{4}\right)^2} =\sqrt{1^2\left(\sqrt{3}\right)^2-4\sqrt{3}+2^2} =\sqrt{1^2\left(\sqrt{3}\right)^2-2\cdot \:1\cdot \:2\sqrt{3}+2^2} =\sqrt{\left(1\cdot \sqrt{3}-2\right)^2} =\sqrt{\left(2-1\cdot \sqrt{3}\right)^2} =2-1\cdot \sqrt{3} =2-\sqrt{3}$

$x+2=2-\sqrt{3}$

$x+2-2=2-\sqrt{3}-2$

$x=-\sqrt{3}$

$x+2=-\sqrt{7-4\sqrt{3}}$

$-\sqrt{7-4\sqrt{3}} =-\left(2-1\cdot \sqrt{3}\right) =-\left(2-\sqrt{3}\right)$

$x+2-2=-\left(2-\sqrt{3}\right)-2$

$x=\sqrt{3}-4$

$x=-\sqrt{3},\:x=\sqrt{3}-4$

Sper că te-am ajutat!

Succes!