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glenn101

Member

Registered: 2008-04-02

Posts: 108

Quadratics and a graphic question

This question is related to quadratics

Make x the subject in each of the following and give the values for t for which real solutions to the equation can be found.

2x^2-4t=x

this is the graphing question.

x-y=5
xy=126

describe in a detailed way, as I am going to be tested on this tomorrow:(
Thanks in advance,
Glenn

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"If your going through hell, keep going."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quadratics and a graphic question

When you have x² and x in an equation, you make x the subject by completing the square.

2x² - 4t = x
2x² - x = 4t
x² - x/2 = 2t
x² - x/2 + 1/16 - 1/16 = 2t
(x - 1/4)² - 1/16 = 2t
(x - 1/4)² = 2t + 1/16
x - 1/4 = ±√(2t + 1/16)
x = 1/4 ±√(2t + 1/16)

A solution only exists if the term inside the square root is non-negative, so from that you can get your condition on t.

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Why did the vector cross the road?
It wanted to be normal.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Quadratics and a graphic question

Isn't this also true?
(x - 1/4)² = 2t + 1/16
+/- (x - 1/4) = +/- √(2t + 1/16)

Last edited by John E. Franklin (2008-04-14 15:28:31)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quadratics and a graphic question

>_<
Fixed, thanks John.

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Why did the vector cross the road?
It wanted to be normal.