Answer:
Let a = side length of a cube
Let S = surface area of a cube
Area of a square = a²
Since a cube has 6 square sides: S = 6a²
To make a the subject:
S = 6a²
Divide both sides by 6:
[tex]\sf \implies \dfrac{S}{6}=a^2[/tex]
Square root both sides:
[tex]\sf \implies a=\sqrt{\dfrac{S}{6}}[/tex]
(positive square root only as distance is positive)
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[tex]\sf x=-3-\sqrt{2} \implies (x+[3+\sqrt{2}])=0[/tex]
[tex]\sf x=-3+\sqrt{2} \implies (x+[3-\sqrt{2}])=0[/tex]
Therefore,
[tex]\sf y=a(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex] for some constant a
Given the y-intercept is at (0, -5)
[tex]\sf \implies a(0+3+\sqrt{2}) (0+3-\sqrt{2})=-5[/tex]
[tex]\sf \implies a(3+\sqrt{2}) (3-\sqrt{2})=-5[/tex]
[tex]\sf \implies a(9-2)=-5[/tex]
[tex]\sf \implies 7a=-5[/tex]
[tex]\sf \implies a=-\dfrac57[/tex]
Substituting found value of a into the equation and simplifying:
[tex]\sf y=-\dfrac57(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]
[tex]\sf \implies y=-\dfrac57(x^2+6x+7)[/tex]
[tex]\sf \implies y=-\dfrac57x^2-\dfrac{30}{7}x-5[/tex]
