Question

$4x^{2} = \frac{9}{32}$
$(x - 3)^{2} = 0.01$
$3 x^{2} - 0.7 = - 0.22$
Dau coroana și 50 puncte:)​

Answer 1

$a) \\4x^2= \frac{9}{32} = > 4x^2 \times 32= 9 = > 128x^2=9 = > x^2= \frac{9}{128} = > \\ x1= \sqrt\frac{9}{128} = \frac{3}{8 \sqrt 2} = \frac{3 \sqrt 2}{16}\\x2= -\sqrt\frac{9}{128}= - \frac{3}{8\sqrt2}= -\frac{3 \sqrt 2}{16}$$b)\\(x-3)^2= 0.01 = > (x-3)= \frac{1}{100} = > x1-3= \sqrt\frac{1}{100} = > x1-3= \frac {1}{10}\\x1=\frac{1}{10}+3= > x1= \frac{31}{10}\\ x2-3= -\sqrt\frac{1}{100}= > x2-3= -\frac{1}{10} = > x2= -\frac{1}{10}+3 = > x2= \frac {29}{10}$

$c)\\3x^2-0.7=-0.22 = > 3x^2= -0.22+0,7= > 3x^2=0,48= > x^2=\frac{0,48}{3}= > x^2= 0,16 = > x^2= \frac{16}{100} = > x^2= \frac{4}{25} \\x1= \sqrt\frac{4}{25}= > x1= \frac{2}{5}\\x2= -\sqrt\frac{4}{25} = > x2= -\frac{2}{5}$