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

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

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Answer 1 · 2016-12-22T20:44:11

$x = 1$ $x=1$ or $x = \frac{1}{2}$ $x=1/2$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We will use this below with $a = (4 x - 3)$ $a=(4x-3)$ and $b = 1$ $b=1$ .

I prefer not to have to do much arithmetic involving fractions, so I would pre-multiply this equation by $8$ $8$ to avoid them and get:

$0 = 8 (2 x^{2} - 3 x + 1)$ $0 = 8(2x^2-3x+1)$

$0 = 16 x^{2} - 24 x + 8$ $color(white)(0) = 16x^2-24x+8$

$0 = 16 x^{2} - 24 x + 9 - 1$ $color(white)(0) = 16x^2-24x+9-1$

$0 = {(4 x - 3)}^{2} - 1^{2}$ $color(white)(0) = (4x-3)^2-1^2$

$0 = ((4 x - 3) - 1) ((4 x - 3) + 1)$ $color(white)(0) = ((4x-3)-1)((4x-3)+1)$

$0 = (4 x - 4) (4 x - 2)$ $color(white)(0) = (4x-4)(4x-2)$

$0 = (4 (x - 1)) (2 (2 x - 1))$ $color(white)(0) = (4(x-1))(2(2x-1))$

$0 = 8 (x - 1) (2 x - 1)$ $color(white)(0) = 8(x-1)(2x-1)$

Hence:

$x = 1$ $x = 1" "$ or $x = \frac{1}{2}$ $" "x = 1/2$

$color(white)()$
Footnote

Why $8$ $8$ ?

$8 = 2 \cdot 2^{2}$ $8 = 2*2^2$

The first factor of $2$ $2$ makes the leading term into a perfect square. The additional $2^{2}$ $2^2$ factor avoids us having to divide $3$ $3$ by $2$ $2$ and end up working with $\frac{1}{2}$ $1/2$ 's and $\frac{1}{4}$ $1/4$ 's

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

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Explanation:

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