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How do you solve $y^{2} - 6 y + 8 = 0$ $y^2-6y+8=0$ ?

3 Answers

Mar 14, 2016

Factor the left side to find
$XXX y = 2 or y = 4$ $color(white)("XXX")y=2 or y=4$

Explanation:

Given
$XXX y^{2} - 6 y + 8 = 0$ $color(white)("XXX")y^2-6y+8=0$

Factoring the left side expression
$XXX (y - 4) (y - 2) = 0$ $color(white)("XXX")(y-4)(y-2)=0$

which implies
either
$XXX (y - 4) = 0 \to y = 4$ $color(white)("XXX")(y-4)=0 rarr y=4$
or
$XXX (y - 2) = 0 \to y = 2$ $color(white)("XXX")(y-2)=0 rarr y=2$

Mar 14, 2016

${2, 4}$ ${2,4}$

Explanation:

$y^{2} - 6 y + 8 = 0$ $y^2-6y+8=0$
$(y - 2) (y - 4) = 0$ $(y-2)(y-4)=0$
$(y - 2) = 0 \Rightarrow y = 2$ $(y-2)=0 => y=2$
$(y - 4) = 0 \Rightarrow y = 4$ $(y-4)=0 =>y=4$
${2, 4}$ ${2,4}$

Mar 14, 2016

$y = 2$ $y=2$ or $y = 4$ $y=4$

Explanation:

Method of factoring:

$y^{2} - 6 y + 8 = 0$ $y^2-6y+8=0$
$(y - 2) (y - 4) = 0$ $(y-2)(y-4)=0$
$y - 2 = 0$ $y-2=0$ or $y - 4 = 0$ $y-4=0$

Solving for $y$ $y$ , gives $y = 2$ $y=2$ or $y = 4$ $y=4$

Method of quadratic equation:

$y = \frac{6 \pm \sqrt{36 - 4 (1) (8)}}{2 (1)} = \frac{6 \pm \sqrt{4}}{2}$ $y=(6+-sqrt(36-4(1)(8)))/(2(1))=(6+-sqrt(4))/(2)$
$y = 2$ $y=2$ or $y = 4$ $y=4$

Method of graphing:
graph{x^2-6x+8 [-0.927, 9.073, -1.44, 3.56]}

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