Not that standard form is y=ax^2+bx+cy=ax2+bx+c
and that vertex form is y=color(red)(a)(x+color(red)(b/(2a)))^2+c +ky=a(x+b2a)2+c+k
where kk is a correction constant in your case a=1a=1
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Given:" "y=x^2+3x-9=0 y=x2+3x−9=0
write as: 0=(x^2+3x)-90=(x2+3x)−9
Add the correction constant kk
0=(x^2+3x)-9+k0=(x2+3x)−9+k
Remove the xx from 3x3x
0=(x^2+3)-9+k0=(x2+3)−9+k
Halve the 3
0=(x^2+3/2)-9+k0=(x2+32)−9+k
Move the power from x^2x2 to outside the bracket
0=(x+3/2)^2-9+k" "color(red)(larr "the error comes from "a(3/(2a))^2)0=(x+32)2−9+k ←the error comes from a(32a)2
but k+(3/2)^2=0 => k=-9/4k+(32)2=0⇒k=−94 giving
color(blue)("Vertex form "->0=(x+3/2)^2- 45/4Vertex form →0=(x+32)2−454
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Take the -45/4−454 to the other side of the equals and change its sign.
(x+3/2)^2=+45/4(x+32)2=+454
Square root both sides
x+3/2=sqrt(45/4)x+32=√454
Move the 3/232 to the other side of =
x=-3/2+-sqrt(5xx9)/sqrt(4)x=−32±√5×9√4
color(blue)(x=-3/2+-(3sqrt(5))/2)x=−32±3√52
color(blue)(x~~1.854" "and" " -4.854 ) x≈1.854 and −4.854
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