How do you solve the quadratic $6 y^{2} - 13 y + 6 = 0$ $6y^2-13y+6=0$ using any method?

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How do you solve the quadratic $6 y^{2} - 13 y + 6 = 0$ $6y^2-13y+6=0$ using any method?

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Answer 1 · 2016-10-06T11:26:29

Using the quadratic formula
$XXX y = \frac{2}{3} or y = \frac{3}{2}$ $color(white)("XXX")color(green)(y=2/3color(black)(" or ")y=3/2)$

Explanation:

The quadratic formula for the general quadratic (in $y$ $y$ )
$XXX a y^{2} + b y + c = 0$ $color(white)("XXX")ay^2+by+c=0$
is
$XXX y = \frac{- b \pm \sqrt{b^{2} + 4 a c}}{2 a}$ $color(white)("XXX")y=(-b+-sqrt(b^2+4ac))/(2a)$

For the example: $6 y^{2} - 13 y + 6 = 0$ $6y^2-13y+6=0$
$XXX a = 6$ $color(white)("XXX")a=6$
$XXX b = - 13$ $color(white)("XXX")b=-13$
$XXX c = 6$ $color(white)("XXX")c=6$
giving
$XXX y = \frac{13 \pm \sqrt{{(- 13)}^{2} - 4 (6) (6)}}{2 (6)}$ $color(white)("XXX")y=(13+-sqrt((-13)^2-4(6)(6)))/(2(6))$

$XXX = \frac{13 \pm \sqrt{169 - 144}}{12}$ $color(white)("XXX")=(13+-sqrt(169-144))/12$

$XXX = \frac{13 \pm 5}{12}$ $color(white)("XXX")=(13+-5)/12$

$XXX = \frac{18}{12} = \frac{3}{2} or = \frac{8}{12} = \frac{2}{3}$ $color(white)("XXX")=18/12= 3/2color(black)(" or ")=8/12=2/3$

Answer 2 · 2016-10-06T11:50:58

$y = \frac{2}{3}$ $y = 2/3$ or $y = \frac{3}{2}$ $y = 3/2$

Explanation:

Use an AC method:

Look for a pair of factors of $A C = 6 \cdot 6 = 36$ $AC = 6*6=36$ with sum $B = 13$ $B=13$

The pair $9, 4$ $9, 4$ works.

Use this pair to split the middle term, then factor by grouping as follows:

$0 = 6 y^{2} - 13 y + 6$ $0 = 6y^2-13y+6$

$0 = 6 y^{2} - 9 y - 4 y + 6$ $color(white)(0) = 6y^2-9y-4y+6$

$0 = (6 y^{2} - 9 y) - (4 y - 6)$ $color(white)(0) = (6y^2-9y)-(4y-6)$

$0 = 3 y (2 y - 3) - 2 (2 y - 3)$ $color(white)(0) = 3y(2y-3)-2(2y-3)$

$0 = (3 y - 2) (2 y - 3)$ $color(white)(0) = (3y-2)(2y-3)$

Hence $y = \frac{2}{3}$ $y = 2/3$ or $y = \frac{3}{2}$ $y = 3/2$

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