Question

Solve the equation 4$x^2$+ 7x = 1 by the method of completing the square. In one of the options the positive root and the converted form of the equation after completing the square is given. Select this option.

• A. x= (7+√65)/8 (2x+7/4)² – 65/4 = 0
• B. x= (-7-√65)/8 (x-7/2)² – 45/4 = 0
• C. x= (3-√19)/2 (5x+3)² – 19 = 0
• D. x= (-7+√65)/8 (2x+7/4)² – 65/16 = 0

Answer 1

The correct option is D

x= (-7+√65)/8 (2x+7/4)² – 65/16 = 0

4$x^2$ + 7x = 1

$(2x{)}^2+2.(\frac{7}{4}\left)\right(2)x=1$

$(2x{)}^2+2.(\frac{7}{4}\left)\right(2)x+(\frac{7}{4}{)}^2=1+(\frac{7}{4}{)}^2$(Add and subtract $(\frac{7}{4}{)}^2$

$(2x+\frac{7}{4}{)}^2$= $\frac{65}{16}$ (Using $(a+b{)}^2$ formula)

$(2x+\frac{7}{4})=\frac{\pm \sqrt{65}}{4}$

$2x=\frac{(-7\pm \sqrt{65})}{4}$

$x=\frac{(-7+\sqrt{65})}{8}or\frac{(-7-\sqrt{6}5)}{8}$