Math, asked by sheaxfonseca, 10 months ago

How do you solve (2x-1)^2 = 0

Answers

Answered by ng580759
1

Step-by-step explanation:

4x^2+1-4x=0 4x^2-4x+1 =0 4x^2-2x-2x+1 2x(2x-1)-1(2x-1)=0 (2x-1)(2x-1)=0 2x-1=0 or 2x-1=0 2x=1 or 2x=1 ×=1/2 or ×=1/2

Answered by Learnaddiction
1

Answer:

The equation we get is 4x^2-4x+1=0..this is by...(a-b)^2=a^2-2ab+b^2

Then by splitting the middle term..

4x^2-2x-2x+1=

2x(2x-1)-1(2x-1)=0

x=1/2

hope u get it

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