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The vertex of the quadratic $x(x-2a)$ occurs when $x=4$.The vertex of the quadratic $(x+a)(x-3a)$ occurs when $x=L$. What is $L$?

EBP
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IMbADdAtMath
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    Please also add your own thoughts and work on the problem – Vizag Jun 14 '19 at 18:09
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    What are the two roots of the equation? Where does the minima/vertex occur? – Vizag Jun 14 '19 at 18:09
  • IMbADdAtMath I see you're relative new to MSE. If you feel that an answer solved the problem, please mark it as 'accepted' by clicking the green check mark $\left(\color{limegreen}{\checkmark}\right)$ . This helps keep the focus on (older) question which has no good answers. – callculus42 Jun 20 '19 at 19:35

2 Answers2

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$x(x- 2a)= x^2- 2ax= x^2- 2ax+ a^2- a^2= (x- a)^2- a^2$. If x= a, that is $-a^2$.0 For any other x it is $-a^2$ plus a positive number so larger than $-a^2$. The vertex is $(a, -a^2)$. We are told that the vertex is at x= 4 so a= 4. Then $(x+a)(x-3a)= (x+4)(x-12)= x^2- 8x- 48= x^2- 8x+ 16- 16+ 48= (x- 4)^2+ 32$. The vertex is again at x= 4.

user247327
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The first coordinate of the vertex is the mean of the zeroes. In the first case that's $(0+2a)/2=a$ and we are told that this value is $4$. In the second case we have $(-a+3a)/2=a$ as well.

Michael Hoppe
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