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In standard form, determine the equation of the quadratic equation that has x-intercepts, √5 and -√5 and contains the point (3,3)?

 

I understand how to get (a).

 

3=a(3+√5)(3-√5)

3=a(9-5)

3=4a

a=3/4

then y=3/4(x+√5)(x-√5)

 

But I'm stuck at where you have to convert it to standard form?

 

I appreciate the help!

 Nov 6, 2016
 #1
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so far so good :)


\( y=3/4(x+\sqrt5)(x-\sqrt5)\\  y=3/4(x^2-5)\\ y=\frac{3}{4}\;x^2\;-\;\frac{15}{4}\)

 Nov 6, 2016

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