solve the quadratic equation by completing the squares:3x^2-2x+1=5

## solve the quadratic equation by completing the squares:3x^2-2x+1=5

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# 4 Answers

3X

^{2}- 2x -4 = 0 That is how quadratic formula has derived. Funding the roots of :

X

^{2 }-2/3 X -4/3 = 0 Factoring by Completing square of: aX

^{2}+ bX + c = 0 X

^{2}-2( 1/3X) + 1/9 = 4/3 + 1/9 X^{2}+ b/a X + c/a = 0 ( X - 1/3 )

^{2 }=13 /9 X^{2 }+ 2 (b/2a) + c/a =0 X

^{2 }+ 2( b/2a) +(b^{2}/ 4a^{2}) = -c/a + (b^{2}/4a^{2}) X = 1/3 ±√13/3 ( X + b/ 2a)

^{2}= ( b^{2}- 4ac) / 4a^{2 } X = - b/2a ±√( b

^{2}- 4ac) /2a In this case a = 3 , b= -2, c = -4

Using quadratic formula will lead the same answer. That is how quadratic formula derived.

Arthur D took "half of 2/3 and square[d] it" to find the constant term of the perfect square trinomial. Why half?

Nataliya D, after multiplying the equation by 3, got (3x)^2 - 2*3x*1 + 1 = 12 + 1. Why 1? (This is a great method to avoid working with fractions!)

Here's "why": (ax + b)^2 = (ax)^2 + 2abx + b^2, or,

if a = 1, (x + b)^2 = x^2 + 2bx + b^2. In completing the square we create an expression like the right side of one of these equations and then substitute the left side for that expression.

Arthur D first divided both sides of the equation by 3 to make a = 1. In the resulting equation he knew, from (x + b)^2 = x^2 + 2bx + b^2, that the coefficient of the linear term had to be 2b, so he halved and squared it to get b^2. Then he added b^2 to
both sides of the equation.

Nataliya D saw the 3x^2 term and decided to use (ax + b)^2 = (ax)^2 + 2abx + b^2 with a = 3. She first multiplied both sides of the equation by a = 3 to get (3x)^2 - 2*3*x - 4*3 = 0. From the pattern she knew that 2ab is the coefficient of the linear term,
so 2*3*b = -2*3 and b = -1.

In both cases once you've found b^2 you have to add it in order to create the perfect square trinomial. Since that changes the value of the equation you must do something to regain the original value. To do that, both tutors added b^2 to the right side
of the equation. Another way, instead of adding just b^2, add (b^2 - b^2). This ensures that you don't forget to regain the original value; esp. in completing the square on two variables in the equations of conic sections.

So for this particular problem I like Nataliya D's method, but I would tweak it a little to make the proper method more apparent.

3x^2-2x+1=5

Subtract 1 from both sides:

3x^2-2x=4

Multiply both sides by 3:

9x^2-6x=12, or

(3x)^2 + 2(-1*3)x = 12

Identify b:

ab = -1 * 3, a = 3, b = -1

Add and subtract b^2:

(3x)^2 + 2(-1*3)x + (-1)^2 - (-1)^2 = 12

The first 3 terms are now a perfect square trinomial; substitute the perfect square binomial:

(3x - 1)^2 - (-1)^2 = 12

Simplify by adding 1 to both sides:

(3x - 1)^2 = 13

Take the square root of both sides (remember that sqrt(k^2) = |k|):

|3x - 1| = sqrt(13)

Simplify:

3x - 1 = +- sqrt(13)

3x = 1 +- sqrt(13)

x = (1 +- sqrt(13))/3

Check: 3x^2 - 2x + 1 =? 5

3((1 +- sqrt(13))/3)^2 - 2((1 +- sqrt(13))/3) + 1

= 3((1 +- 2sqrt(13) +13)/(3*3)) - 2((1 +- sqrt(13))/3) + 1

= ((1 +- 2sqrt(13) +13)/3) - 2((1 +- sqrt(13))/3) + 1

= 14/3 + (2/3)*(+- sqrt(13)) - (2/3) - (2/3)*(+- sqrt(13)) + 1

= 12/3 + 1

= 5 √

3x^2-2x+1=5

3x^2-2x=5-1

3x^2-2x=4

(3x^2)/3-(2x)/3=4/3

x^2-(2/3)x=4/3

take half of 2/3 and square it

x^2-(2/3)x+(1/9)=(4/3)+(1/9)

[x-(1/3)]^2=(13/9)

take the square root of both sides

x-(1/3)=+sqrt[(13/9)]

x=(1/3)+[sqrt(13)/sqrt(9)]

x=(1/3)+sqrt(13)/3

x=[1+sqrt(13)]/3

3x

3x

9x

9x

(3x)

(3x - 1)

3x - 1 = ±√13

3x = 1 ± √13

^{2}- 2x + 1 = 53x

^{2}- 2x - 4 = 0 . Let's multiply this equation by 39x

^{2}- 6x - 12 = 09x

^{2}- 6x = 12(3x)

^{2}- 2*3x*1 + 1 = 12 + 1(3x - 1)

^{2}= 133x - 1 = ±√13

3x = 1 ± √13

*1 ± √13**x = -----------**3*