showing that one equation satisfy an other one

[petr]

Hi,

i am doing past papers and i got stuck by question 6) a) i)

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6 . . .The diagram shows a curve and a line which intersect at points A, B, and C.

The curve has equation y = x³ - x² - 5x + 7, and the straight line has equation y = x + 7. The point B has coordinates (0, 7).

(a) (i) . . .Show that the x -coordinate of the points A and C satisfy the equation x² - x - 6 = 0.

http://filestore.aqa.org.uk/subjects/AQA-MPC1-QP-JUN14.PDF

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x³ - x² - 5x + 7 = x + 7

x³ - x² - 5x = x

x not equal to 0, so dividing through gives

x² - x - 5 = 1

x² - x - 6 = 0

http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-MS-JUN14.PDF

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x^3-x^2-5x+7 = x+7


they want me to show that the x-coordinates of the points A and C satisfy the equation (x^2-x-6=0)...

so
x^3-x^2-5x+7 = x+7
=> x^3-x^2-6x = 0
and at this point it looks like they take dy/dx and got this>
x^2-x-6=0

which is correct colution and my question is why?

please help@. I cannot figure it out.

 

[Ishuda]

Hi,

i am doing past papers and i got stuck by question 6) i) a)

http://filestore.aqa.org.uk/subjects/AQA-MPC1-QP-JUN14.PDF

the solution is here> 6) i) a)
http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-MS-JUN14.PDF


x^3-x^2-5x+7 = x+7


they want me to show that the x-coordinates of the points A and C satisfy the equation (x^2-x-6=0)...

so
x^3-x^2-5x+7 = x+7
=> x^3-x^2-6x = 0
and at this point it looks like they take dy/dx and got this>
x^2-x-6=0

which is correct colution and my question is why?

please help@. I cannot figure it out.

No, what they do is factor the expression
x³ - x² - 6 x = x (x² - x - 6)
So the equation
x³ - x² - 6 x = 0
has a solution of
x = 0
and
x² - x - 6 = 0
which one can solve via the quadratic formula.