# sketching curves, roots problem

• Oct 24th 2010, 12:36 PM
chartsy
sketching curves, roots problem
heres the question: sketch the curve of y=4-7x-2xsqrd
i get y=4, but to get the two x roots i factorise as normal, and get the solutions x=1/2 and x=+4
however the textbook says that the second root is actually -4...

why?
• Oct 24th 2010, 12:45 PM
mr fantastic
Quote:

Originally Posted by chartsy
heres the question: sketch the curve of y=4-7x-2xsqrd
i get y=4, but to get the two x roots i factorise as normal, and get the solutions x=1/2 and x=+4
however the textbook says that the second root is actually -4...

why?

• Oct 24th 2010, 01:44 PM
chartsy
for easiness i will make this smiley face the sign for squared:(Nod)
sketch curve of y=4-7x-2x(Nod)

1) a<0 (-2) so curve is inverted u shape
2) find where it crosses y axis:
subst x=o into eqn
y=4
3)factorise to find the two roots of the curve
0=-2x(Nod)-7x+4
0=-2x(Nod)-8x+x+4

-2x(x-4) 1(x-4)
(-2x+1) (x-4)=0

1)x-4=0
x=4
2)-2x+1=0
x=1/2

so x=1/2 and x=4

so why is it -4?
• Oct 24th 2010, 06:20 PM
skeeter
Quote:

Originally Posted by chartsy
1) a<0 (-2) so curve is inverted u shape
2) find where it crosses y axis:
subst x=o into eqn
y=4
3)factorise to find the two roots of the curve
0=-2x^2-7x+4
0=-2x^2-8x+x+4

-2x(x-4) 1(x-4) ... should be -2x(x+4) + 1(x+4)
(-2x+1) (x-4)=0

1)x-4=0
x=4
2)-2x+1=0
x=1/2

so x=1/2 and x=4

so why is it -4?

$4 - 7x - 2x^2 = 0$

$(4 + x)(1 - 2x) = 0$

$4 + x = 0$ ... $x = -4$

$1 - 2x = 0$ ... $x = 1/2$

next time, use the caret (^) to indicate an exponent.
• Oct 25th 2010, 12:08 AM
chartsy
hmmm thanks for that....need to be more focused when factorising then
• Oct 25th 2010, 04:36 AM
mr fantastic
Quote:

Originally Posted by chartsy
hmmm thanks for that....need to be more focused when factorising then

If you had checked it by expanding, you would have seen the mistake immediately.