How do you solve the quadratic equation by completing the square: $y^{2} + 3 y = 10$ $y^2+3y=10$ ?

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How do you solve the quadratic equation by completing the square: $y^{2} + 3 y = 10$ $y^2+3y=10$ ?

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Answer 1 · 2018-04-22T16:26:09

$y = 2 or y = - 5$ $y=2 or y=-5$

Explanation:

$y^{2} + 3 y = 10$ $y^2 +3y" "= 10$

You want to add in a missing term on the left so that you have a perfect square trinomial which can then be written in the form
${(y + ?)}^{2}$ $(y +?)^2$ This process is called 'completing the square.'

The 'missing term' is found from ${(\frac{b}{2})}^{2}$ $(b/2)^2$ and has to be added to both sides of the equation.

$y^{2} + 3 y + {(\frac{3}{2})}^{2} = 10 + {(\frac{3}{2})}^{2}$ $y^2 +3y" "color(blue)( +(3/2)^2) = 10" " color(blue)(+(3/2)^2)$

The left side can now be written as ${(y + ?)}^{2}$ $(y+?)^2$ and the right side can be simplified.

${(y + \frac{3}{2})}^{2} = 10 + \frac{9}{4} \leftarrow$ $(y +3/2)^2 = 10 +9/4" "larr$ square root both sides

$(y + \frac{3}{2}) = \pm \sqrt{10 + 2 \frac{1}{4}} = \pm \sqrt{12 \frac{1}{4}}$ $(y+3/2) = +-sqrt(10 +2 1/4) = +-sqrt(12 1/4)$

$y + \frac{3}{2} = \pm \sqrt{\frac{49}{4}}$ $y +3/2 = +-sqrt(49/4)$

$y = \frac{7}{2} - \frac{3}{2} or y = - \frac{7}{2} - \frac{3}{2}$ $y = 7/2-3/2" "or" "y = -7/2-3/2$

$y = \frac{4}{2} = 2 or y = - \frac{10}{2} = - 5$ $y = 4/2=2" "or " "y= -10/2=-5$

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How do you solve the quadratic equation by completing the square: $y^{2} + 3 y = 10$ $y^2+3y=10$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the quadratic equation by completing the square: y2+3y=10y^2+3y=10?

1 Answer

Explanation:

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How do you solve the quadratic equation by completing the square: $y^{2} + 3 y = 10$ $y^2+3y=10$ ?