How do you solve $2 x^{2} + 5 x - 3 = 0$ $2x^2 + 5x -3=0$ by completing the square?

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How do you solve $2 x^{2} + 5 x - 3 = 0$ $2x^2 + 5x -3=0$ by completing the square?

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Answer 1 · 2016-04-25T20:31:58

$x = \frac{1}{2}$ $x = 1/2$ or $x = - 3$ $x=-3$

Explanation:

We will use the difference of squares identity, which can be written:

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

First, to reduce the amount of arithmetic involving fractions, first multiply through by $2^{3} = 8$ $2^3 = 8$ to get:

$0 = 16 x^{2} + 40 x - 24$ $0 = 16x^2+40x-24$

$= {(4 x)}^{2} + 2 (5) (4 x) - 24$ $= (4x)^2+2(5)(4x)-24$

$= {(4 x + 5)}^{2} - 25 - 24$ $= (4x+5)^2-25-24$

$= {(4 x + 5)}^{2} - 7^{2}$ $=(4x+5)^2-7^2$

$= ((4 x + 5) - 7) ((4 x + 5) + 7)$ $=((4x+5)-7)((4x+5)+7)$

$= (4 x - 2) (4 x + 12)$ $=(4x-2)(4x+12)$

$= (2 (2 x - 1)) (4 (x + 3))$ $=(2(2x-1))(4(x+3))$

$= 8 (2 x - 1) (x + 3)$ $=8(2x-1)(x+3)$

Hence:

$x = \frac{1}{2}$ $x = 1/2$ or $x = - 3$ $x=-3$

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Why did I premultiply by $8$ $8$ ?

One factor of $2$ $2$ makes the leading term into a perfect square, then the additional factor $2^{2} = 4$ $2^2 = 4$ compensates for the middle term being an odd number.

Otherwise, we might proceed as follows:

$0 = 2 x^{2} + 5 x - 3$ $0 = 2x^2+5x-3$

$= 2 (x^{2} + \frac{5}{2} x - \frac{3}{2})$ $=2(x^2+5/2x-3/2)$

$= 2 ({(x + \frac{5}{4})}^{2} - \frac{25}{16} - \frac{3}{2})$ $=2((x+5/4)^2-25/16-3/2)$

$= 2 ({(x + \frac{5}{4})}^{2} - \frac{49}{16})$ $=2((x+5/4)^2-49/16)$

$= 2 ({(x + \frac{5}{4})}^{2} - \frac{7^{2}}{4^{2}})$ $=2((x+5/4)^2-7^2/4^2)$

$= 2 ((x + \frac{5}{4}) - \frac{7}{4}) ((x + \frac{5}{4}) + \frac{7}{4})$ $=2((x+5/4)-7/4)((x+5/4)+7/4)$

$= 2 (x - \frac{2}{4}) (x + \frac{12}{4})$ $=2(x-2/4)(x+12/4)$

$= 2 (x - \frac{1}{2}) (x + 3)$ $=2(x-1/2)(x+3)$

Hence:

$x = \frac{1}{2}$ $x = 1/2$ or $x = - 3$ $x=-3$

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