Respuesta :

Answer:

[tex]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[/tex]

Leora03

Answer:

x= [tex]\frac{1}{2}[/tex]     or     x= -3

Step-by-step explanation:

[tex]\boxed{x^{2} +kx=(x+\frac{k}{2})^{2} -(\frac{k}{2})^{2} }[/tex]

First ensure that the coefficient of x² is 1.

x² +[tex]\frac{5}{2}[/tex]x -[tex]\frac{3}{2}[/tex]= 0

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

(x +[tex]\frac{5}{4}[/tex])² -([tex]\frac{5}{4}[/tex])² -[tex]\frac{3}{2}[/tex]= 0

(x +[tex]\frac{5}{4}[/tex])²- [tex]\frac{25}{16}[/tex] -[tex]\frac{3}{2}[/tex]= 0

(x +[tex]\frac{5}{4}[/tex])² -[tex]\frac{49}{16}[/tex]= 0

(x +[tex]\frac{5}{4}[/tex])²= [tex]\frac{49}{16}[/tex]

x +[tex]\frac{5}{4}[/tex]= [tex]\sqrt{\frac{49}{16} }[/tex]                (square root both sides)

x +[tex]\frac{5}{4}[/tex]= ±[tex]\frac{7}{4}[/tex]

x= -[tex]\frac{5}{4}[/tex] +[tex]\frac{7}{4}[/tex]       or       x= -[tex]\frac{5}{4}[/tex] -[tex]\frac{7}{4}[/tex]

x= [tex]\frac{1}{2}[/tex]             or       x= -3

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