3 bài tập Giải phương trình bậc hai dạng đặc biệt (có lời giải)

a) ${\rm{ }}{x^2} - 8 = 0 \Leftrightarrow {x^2} - 8 \Leftrightarrow x = \pm 2\sqrt 2$

Phương trình có hai nghiệm : ${x_1} = - 2\sqrt 2 :{x_2} = 2\sqrt 2$

b) $5{x^2} - 20 = 0 \Leftrightarrow {x^2} = 4 \Leftrightarrow x = \pm 2$

Tập nghiệm $S = \left\{ { - 2;{\rm{ }}2} \right\}$

c) Vì $0{,}4{x^2} + 1{\rm{ }} > {\rm{ }}0$ với mọi x nên phương trình $0{,}4{x^2} + 1{\rm{ }} = {\rm{ }}0$ vô nghiệm

d) $2{x^2} + \sqrt 2 x = 0 \Leftrightarrow x(2x + \sqrt 2 ) = 0 \Leftrightarrow \left[ {\begin{array}{llllllllllllllllllll}{x = 0}\\{2x + \sqrt 2 = 0}\end{array}} \right.$$\Leftrightarrow \left[ {\begin{array}{llllllllllllllllllll}{x = 0}\\{x = - \frac{{\sqrt 2 }}{2}}\end{array}} \right.$

Tập nghiệm $S = \left\{ {0; - \frac{{\sqrt 2 }}{2}} \right\}$

${\rm{ e) }} - 0{,}4{x^2} + 1{,}2x = 0 \Leftrightarrow x( - 0{,}4x + 1{,}2) = 0 \Leftrightarrow \left[ {\begin{array}{llllllllllllllllllll}{x = 0}\\{ - 0{,}4x + 1{,}2 = 0}\end{array}} \right.$$\Leftrightarrow \left[ {\begin{array}{llllllllllllllllllll}{x = 0}\\{x = 3}\end{array}} \right.$

Vậy tập nghiệm phương trình là : $S = \left\{ {0;{\rm{ }}3} \right\}$

a) Ta có $6{x^2} + 8x = - 2$

$${x^2} + 8x + 16 = 14$$

$${(x + 4)^2} = 14$$

$$\left[ {\begin{array}{llllllllllllllllllll}{x + 4 = \sqrt {14} }\\{x + 4 = - \sqrt {14} }\end{array}} \right.$$

$$\left[ {\begin{array}{llllllllllllllllllll}{x = - 4 + \sqrt {14} }\\{x = - 4 - \sqrt {14} }\end{array}} \right.$$

b) ${x^2} + 2x = \frac{1}{3}$

${x^2} + 2x = 1 = \frac{4}{3}$

${(x + 1)^2} = \frac{4}{3}$

$\left[ {\begin{array}{llllllllllllllllllll}{x + 1 = \frac{2}{{\sqrt 3 }}}\\{x + 1 = - \frac{2}{{\sqrt 3 }}}\end{array}} \right.$

$\left[ {\begin{array}{llllllllllllllllllll}{x = \frac{2}{{\sqrt 3 }} - 1}\\{x = - \frac{2}{{\sqrt 3 }} - 1}\end{array}} \right.$

$2{x^2} + 5x + 2 = 0$

$2{x^2} + 5x = - 2$

$$2\left( {{x^2} + \frac{5}{2}x + \frac{{25}}{{16}}} \right) = - 2 + \frac{{25}}{8}$$

$2{\left( {x + \frac{5}{4}} \right)^2} = \frac{9}{8}$ $\Leftrightarrow {\left( {x + \frac{5}{4}} \right)^2} = \frac{9}{{16}}$

$\left[ {\begin{array}{llllllllllllllllllll}{x + \frac{5}{4} = \frac{3}{4}}\\{x + \frac{5}{4} = - \frac{3}{4}}\end{array}} \right.$

$\left[ {\begin{array}{llllllllllllllllllll}{x = - \frac{1}{2}}\\{x = - 2}\end{array}} \right.$

Vậy phương trình có hai nghiệm ${x_1} = - 2;{x_2} = - \frac{1}{2}$