Question

solve the following equation by completing square 7x square -2x-1=0​

Answer 1

Answer:

$7x {}^{2} - 2x - 1 = 0 \\ = 7x {}^{2} - 2x = 1 \\ = divide \: both \: sides \: by \: 7 \\ = {x}^{2} - \frac{2x}{7} = \frac{1}{7} \\ add \: ( \frac{2}{14} ) {}^{2} \: to \: both \: sides \\ {x}^{2} - \frac{2x}{7} + ( \frac{2}{14} ) {}^{2} = \frac{1}{7} + ( \frac{2}{14} ) {}^{2} \\ = (x - \frac{2}{14} ) {}^{2} = \frac{1}{7} + \frac{4}{196} \\ = (x - \frac{2}{14} ) {}^{2} = \frac{8}{49} \\ = x - \frac{1}{7} = \sqrt{ \frac{8}{49} } \\ = x - \frac{1}{7} = \frac{ 2\sqrt{2} }{7} \\ = x = \frac{2 \sqrt{2} }{7} + \frac{1}{7} \\ = x = \frac{1 + 2 \sqrt{2} }{7} \\ we \: know \: that \: x {}^{2} \: has \: two \: values \: one \: in \: ( - ) \: and \: other \: is \: in \: positive \: so \: \\ final \: answer \: will \: be \: \\ x = (- \frac{1 + 2 \sqrt{2} }{7} ) \: \: and \: ( + \frac{1 + 2 \sqrt{2} }{7} ) \: \:$

Jai Hind ❤️.

Answer 2

solution:-

=> 7x²-2x-1=0

Solving 7x²-2x-1 = 0 by Completing The Square .

x²-(2/7)x-(1/7) = 0

x²-(2/7)x = 1/7

add both side 1/49

=> x²-(2/7)x+(1/49) = 8/49 ...............(1)

=> x²-(2/7)x+(1/49) = (x-(1/7))²..........(2)

from ( 1 ) and ( 2 )

(x-(1/7))² = 8/49

x = 1/7 + √ 8/49

has two solutions:

x = 1/7 + √ 8/49

or

x = 1/7 - √ 8/49

Note that √ 8/49 can be written as

√ 8 / √ 49 which is √ 8 / 7

I hope it helps you.....