Question

2x²+5x-3=0using completing the square method​

Answer 1

Answer:

x= $\frac{1}{2}$     or     x= -3

Step-by-step explanation:

$\boxed{x^{2} +kx=(x+\frac{k}{2})^{2} -(\frac{k}{2})^{2} }$

First ensure that the coefficient of x² is 1.

x² +$\frac{5}{2}$x -$\frac{3}{2}$= 0

[x +($\frac{5}{2}$ ÷2)]² -($\frac{5}{2}$ ÷2)² -$\frac{3}{2}$= 0

(x +$\frac{5}{4}$)² -($\frac{5}{4}$)² -$\frac{3}{2}$= 0

(x +$\frac{5}{4}$)²- $\frac{25}{16}$ -$\frac{3}{2}$= 0

(x +$\frac{5}{4}$)² -$\frac{49}{16}$= 0

(x +$\frac{5}{4}$)²= $\frac{49}{16}$

x +$\frac{5}{4}$= $\sqrt{\frac{49}{16} }$                (square root both sides)

x +$\frac{5}{4}$= ±$\frac{7}{4}$

x= -$\frac{5}{4}$ +$\frac{7}{4}$       or       x= -$\frac{5}{4}$ -$\frac{7}{4}$

x= $\frac{1}{2}$             or       x= -3

Answer 2

Answer:

$2 {x}^{2} + 5x - 3 = 0 \\ 2( {x}^{2} + \frac{5}{2} x - \frac{3}{2} ) = 0 \\ 2( {x}^{2} + \frac{5}{2} x + {( \frac{5}{4} )}^{2} ) - \frac{3}{2} - {( \frac{5}{4} )}^{2} ) = 0 \\ ( {(x + \frac{5}{4} )}^{2} = \frac{49}{16} \\ x + \frac{5}{4} = ± \frac{7}{4} \\ x = 0.5 \: \: and \: \: 3$