Question

1) 2х^2-х=0
2) 8х^2-2=0
3) 9х^2-6х+1=0
4) 2х^2-х+3=0
5) 2х^2-х-6=0
????????????????​

Answer 1

1) 2x^2-x=0

$\frac{-\left(-1\right)+\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}$

$=\frac{1+\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}$

$= 1 + \sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}$

$= \sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}$

$\left(-1\right)^2=1$

$\left(-1\right)^2$ = 1

$4\cdot \:2\cdot \:0$ = 0

$=\sqrt{1-0}$

$=\sqrt{1}$ = 1

=1+1 = 2

$=\frac{2}{2\cdot \:2}$

$=\frac{2}{4}$

$=\frac{1}{2}$

$\frac{-\left(-1\right)-\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}$

$=\frac{1-\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}$

$1-\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}$

$\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:0}$

$\left(-1\right)^2=1$

$\left(-1\right)^2=1$

$4\cdot \:2\cdot \:0=0$

$= \sqrt{1-0} = \sqrt{1} = 1$

2) 8х^2-2=0

$8x^2-2+2=0+2$

$8x^2=2$

$\frac{8x^2}{8}=\frac{2}{8}$

$x^2=\frac{1}{4}$

$x=\sqrt{\frac{1}{4}}$ = $\frac{1}{2}$

$x=\sqrt{\frac{1}{4}}$ = $\frac{1}{2}$

3) $9x^2-6x+1=0$

$\left(-6\right)^2-4\cdot \:9\cdot \:1$

$=6^2-36$

=36-36 = 0

$x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:9}$

$x=\frac{-\left(-6\right)}{2\cdot \:9}$

$\frac{-\left(-6\right)}{2\cdot \:9}$ = $\frac{6}{2\cdot \:9}$ $=\frac{6}{18}$ $=\frac{1}{3}$

4) 2х^2-х+3=0

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

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

$\left(-1\right)^2=1$ $\left(-1\right)^2$

$4\cdot \:2\cdot \:3=24$

$=\frac{1+\sqrt{23}i}{2\cdot \:2}$

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

$1-\sqrt{\left(-1\right)^2-4\cdot \:2\cdot \:3}=1-\sqrt{23}i$

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

$=1-\sqrt{23}i$

$=\frac{1}{4}-\frac{\sqrt{23}}{4}i$

$x=\frac{1}{4}+i\frac{\sqrt{23}}{4},\:x=\frac{1}{4}-i\frac{\sqrt{23}}{4}$

5) 2х^2-х-6=0

$\frac{-\left(-1\right)+\sqrt{\left(-1\right)^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}$

$=\frac{1+\sqrt{\left(-1\right)^2+4\cdot \:2\cdot \:6}}{2\cdot \:2}$

$1+\sqrt{\left(-1\right)^2+4\cdot \:2\cdot \:6}=1+\sqrt{49}$

$1+\sqrt{\left(-1\right)^2+4\cdot \:2\cdot \:6}$

$\sqrt{\left(-1\right)^2+4\cdot \:2\cdot \:6}$

$\left(-1\right)^2=1$ $1^2$ = 1

$=\sqrt{1+48}$

$\frac{1+\sqrt{49}}{2\cdot \:2}$

$=\frac{1+\sqrt{49}}{4}$

$\sqrt{49}=7$

$=\frac{1+7}{4}$

$=\frac{8}{4}$

=2

$x=\frac{-\left(-1\right)-\sqrt{\left(-1\right)^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}$

$\frac{-\left(-1\right)-\sqrt{\left(-1\right)^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}$

$=\frac{1-\sqrt{49}}{2\cdot \:2}$

$=\frac{1-\sqrt{49}}{4}$

$\sqrt{49}=7$

$=\frac{1-7}{4}$

$=\frac{-6}{4}$$=-\frac{6}{4}$

$=-\frac{3}{2}$

$x=2,\:x=-\frac{3}{2}$

Sper că te-am ajutat!

Succes!