# Thread: sketching curves, roots problem

1. ## 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?

2. 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?

3. for easiness i will make this smiley face the sign for squaredNod)
sketch curve of y=4-7x-2x

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-7x+4
0=-2x-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?

4. 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?
$\displaystyle 4 - 7x - 2x^2 = 0$

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

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

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

next time, use the caret (^) to indicate an exponent.

5. hmmm thanks for that....need to be more focused when factorising then

6. 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.