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Question: How do you solve \(2{{x}^{2}}-3x+1=0\) by completing the square?...

How do you solve 2x23x+1=02{{x}^{2}}-3x+1=0 by completing the square?

Explanation

Solution

To avoid arithmetic multiplying fraction multiply the given equation 2x23x+1=02{{x}^{2}}-3x+1=0 by 8'8'
Use the difference of square identity that is a2b2=(ab)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
For simplifying the problem replace x'x' terms by a'a' and b'b'
Place x'x' term on the left side to find the value of x'x' and the coefficient on the right side.

Complete step by step solution: As per the given equation is 2x23x+1=02{{x}^{2}}-3x+1=0
Here to avoid arithmetic involving fraction multiply the whole equation by 8'8'
8(2x23x+1)=0×88\left( 2{{x}^{2}}-3x+1 \right)=0\times 8
So, after multiplying the modified equation will be.
16x224x+8=016{{x}^{2}}-24x+8=0
Here you can write +91'+9-1'on the place of +8'+8'
16x224x+91=016{{x}^{2}}-24x+9-1=0
Here in above equation, 16x224x+916{{x}^{2}}-24x+9
(ab)2=a22ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} formula
Here in 16x224x+916{{x}^{2}}-24x+9
a=4xa=4x
And b=3b=3
Therefore, process this step:
(4x3)21=0{{\left( 4x-3 \right)}^{2}}-1=0
Here, 1'1' can be written as 12'{{1}^{2}}' because 12=1{{1}^{2}}=1
So,
(4x3)212=0{{\left( 4x-3 \right)}^{2}}-{{1}^{2}}=0
Now in above equation you can apply a2b2=(ab)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) identity.
Here, a=4x3a=4x-3
b=1b=1
Therefore,
[(4x3)1][(4x3)+1]=0\left[ \left( 4x-3 \right)-1 \right]\left[ \left( 4x-3 \right)+1 \right]=0
Opening, the bracket to solve the equation or simplifying the equation.
[4x31][4x3+1]=0\left[ 4x-3-1 \right]\left[ 4x-3+1 \right]=0
(4x4)(4x2)=0\left( 4x-4 \right)\left( 4x-2 \right)=0
Choose common term from each equation in (4x4)\left( 4x-4 \right) 4'4' is common term as it divides by both 4x'4x' and 4'4' and in 2nd{{2}^{nd}} equation (4x2)2\left( 4x-2 \right)'2' is common term as it divides by 4x'4x' and 2'2' both.
So,
[4(x1)][2(2x1)]=0\left[ 4\left( x-1 \right) \right]\left[ 2\left( 2x-1 \right) \right]=0
Multiply 4×2=84\times 2=8
8(x1)(2x1)=08\left( x-1 \right)\left( 2x-1 \right)=0
For finding the value of x,x,
8(x1)=08\left( x-1 \right)=0 or 8(2x1)=08\left( 2x-1 \right)=0
8x8=08x-8=0 or 16x8=016x-8=0
x=88x=\dfrac{8}{8} or x=816x=\dfrac{8}{16}
x=1x=1 and x=12x=\dfrac{1}{2}
The value of xx is 11 and 12\dfrac{1}{2}

Additional Information:
In order to use the completing the square method, the value of aa in quadratic equation must be 1.1. If it is not 1,1, you will have to use AC method or the quadratic formula in order to solve for x.x.
Completing the square is a method used to solve quadratic equation ax2+bx+ca{{x}^{2}}+bx+c where a must be 1.1. The goal is to force a perfect square trinomial on one side and then solve for x'x' by taking the square root of both sides.

Note:
Use the difference of square identity that is a2b2=(ab)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) make the equation into difference of square identity.
As you have multiplied the given equation by 8'8' to avoid too much arithmetic involving fraction.
The digit 8'8' is preferred to multiply because 8=2.228={{2.2}^{2}} the first factor of 22makes the leading term into a perfect square. The additional 22{{2}^{2}} factor avoids us having to divide, 33 by 22 and end up working with 12\dfrac{1}{2} and 14\dfrac{1}{4}