Question

How to solve 7x2 + 2x - 1 = 0 by completing square method

Answer 1

7x²+2x-1=0

the coefficient of x² should be one if not multiply the coefficient of x² with all the numbers

7x²/7 +2x/7 -1/7 =0

x²+2x/7-1/7=0

x²+2x/7=0+1/7

now multiply the coefficient of x with 1/2 and square it

x²+2x/7+(1/7)²=1/7+(1/7)²

(x+1/7)²=1/7+1/49

take LCM

= 7×1+1×1/49

=7+1/49

(x+1/7)² =8/49

√(x+1/7)²= √8/49

cancel the root and square of x+1/7

x+1/7= + - 2√2/7

x= -1 +- 2√2/7

hope it helps you

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Answer 2

Answer:

Step-by-step explanation:

Given: 7x²+2x-1=0

To  find: Solution of equation using completing square method.

Solution:

Step 1:Take the constant to other side

7x²+2x=1

Step 2:Divide the equation by 7

$\frac{7x^{2}+2x }{7} =\frac{1}{7} \\ \\ x^{2} +\frac{2x}{7} =\frac{1}{7}$

Step 3: Manuplute the middle, so that both a and b can be identify from the identity (a+b)²=a²+2ab+b²

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

it is clear that a=x and b=1/7

Step 4: Add 1/49 both sides, square of 1/7

$(x)^{2} +2.x.\frac{1}{7}+\frac{1}{49} =\frac{1}{7}+\frac{1}{49}$

or

$(x+\frac{1}{7} )^{2} =\frac{8}{49}$

or

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

Step 5: Taking square root both sides

$(x+\frac{1}{7} )=\pm\left(\frac{2\sqrt{2} }{7}\right)$

Step 6: Taking different sign find both values of x

$(x+\frac{1}{7} )=\left(\frac{2\sqrt{2} }{7}\right)\\\\x=\frac{2\sqrt{2}}{7}-\frac{1}{7}\\ \\ x=\frac{2\sqrt{2}-1 }{7}$

By the same way

$(x+\frac{1}{7} )=-\left(\frac{2\sqrt{2} }{7}\right)\\\\x=\frac{-2\sqrt{2}}{7}-\frac{1}{7}\\ \\ x=\frac{-2\sqrt{2}-1 }{7}$

Final Answer:

$\bold{x=\frac{2\sqrt{2}-1 }{7}}\\ \\ \bold{x=\frac{-2\sqrt{2}-1 }{7}}$

Hope it helps you.

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