Question

Solve the equation 4$x^2$+ 7x = 1 by the method of completing the square. In one of the options, one of the roots 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)

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

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

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