Question

Solve the quadratic equation 4$x^2$+ 5$x$ = 0 by completion of squares method. In the given options one of the options has one root and also the converted form of equation after completing the square. Select this option

• A. x= 0, $({2x+\frac{5}{4})}^2$ - $\frac{5}{2}$ =0
• B. $x$= $-\frac{5}{4}$, $({2x+\frac{5}{4})}^2$ - $\frac{5}{2}$ =0
• C. $x$= $-\frac{5}{4}$, $({2x+\frac{5}{4})}^2$ -$({\frac{5}{4})}^2$=0
• D. $x$= $-\frac{5}{2}$ , $(2x-5{)}^2$ - $({\frac{5}{4})}^2$=0

Answer 1

The correct option is C

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

4$x^2$+ 5$x$ = 0

2$x^2$ + 2.($\frac{5}{4}$)(2)$x$ = 0

2$x^2$+ 2.($\frac{5}{4}$)(2)$x$ +${\left(\frac{5}{4}\right)}^2$ = ${\left(\frac{5}{4}\right)}^2$ (Adding ${\left(\frac{5}{4}\right)}^2$ to both sides of equation)

$(2x+\left(\frac{5}{4}\right){)}^2$= ${\left(\frac{5}{4}\right)}^2$ (Using ${(a+b)}^2$formula)

(2$x$ + $\frac{5}{4}$ ) = $\pm$$\frac{5}{4}$
(Taking square root on both sides)

2$x$ = 0,-$\frac{5}{2}$

$x$= 0, -$\frac{5}{4}$