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

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

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Answer 1 · 2016-04-29T16:55:30

$y = \frac{5}{2} \pm \frac{\sqrt{37}}{2}$ $y = 5/2+-sqrt(37)/2$

Explanation:

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

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

with $a = (2 y - 5)$ $a=(2y-5)$ and $b = \sqrt{37}$ $b=sqrt(37)$ .

Pre-multiply by $4$ $4$ to reduce the amount of working in fractions:

$0 = 4 (y^{2} - 5 y - 3)$ $0 = 4(y^2-5y-3)$

$= 4 y^{2} - 20 y - 12$ $=4y^2-20y-12$

$= {(2 y - 5)}^{2} - 25 - 12$ $=(2y-5)^2-25-12$

$= {(2 y - 5)}^{2} - {(\sqrt{37})}^{2}$ $=(2y-5)^2-(sqrt(37))^2$

$= ((2 y - 5) - \sqrt{37}) ((2 y - 5) + \sqrt{37})$ $=((2y-5)-sqrt(37))((2y-5)+sqrt(37))$

$= (2 y - 5 - \sqrt{37}) (2 y - 5 + \sqrt{37})$ $=(2y-5-sqrt(37))(2y-5+sqrt(37))$

Hence:

$y = \frac{5}{2} \pm \frac{\sqrt{37}}{2}$ $y = 5/2+-sqrt(37)/2$

Answer 2 · 2016-05-02T01:38:31

$y = 5.541$ $y = 5.541$ OR $y = - 0.541$ $y = -0.541$

Explanation:

$y^{2} - 5 y = 3 \Rightarrow$ $y^2 - 5y = 3 rArr$ move the constant to the right hand side.

Complete the square by adding what is missing from the square of the binomial to BOTH sides. ${(b \div 2)}^{2}$ $(b÷2)^2$

$y^{2} - 5 y + {\frac{5}{2}}^{2} = 3 + {\frac{5}{2}}^{2}$ $y^2 - 5y + color(red) (5/2)^2 = 3 + color(red)(5/2)^2$

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

$y - \frac{5}{2} = \pm \sqrt{9.25}$ $y - 5/2 = +-sqrt(9.25$ ............ $\Rightarrow 3 + 6 \frac{1}{4} = 9 \frac{1}{4}$ $rArr 3+6 1/4 = 9 1/4$

$y = \sqrt{9.25} + 2.5$ $y = sqrt9.25 +2.5$ OR $y = - \sqrt{9.25} + 2.5$ $y = -sqrt9.25 +2.5$

$y = 5.541$ $y = 5.541$ OR $y = - 0.541$ $y = -0.541$

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

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