#1
Okay i'm just doing a bit of maths revision and i'm slightly confused by the steps i need to go through to complete a quadratic equation

Q= Solve the following quadratic inequalities by first sketching appropriate graphs

a) (x+1) (x-4) < 0
= X[squared] - 3x - 4 < 0

So right off the bat you know that the X intercepts of the curve would be at -1 and 4

So thats listed, but how am i supposed to find the vertex? I've only seen how to do it with completed square equations
#3
Nope, not yet, i will soon, but we're supposed to do this without the calcs and show the working
#5
crosses y axis when x is 0, so at -4
Then you draw a graph of a parabola going through the axis at those points and see which bits are <0
Theres your answer -1<x<4
Last edited by add666 at Oct 11, 2007,
#6
convert it to a completed square equation.
(x-1.5)^2

that should start you off. multiply out the brackets and find what you need to add to make it the same as (x+1)(x-4)
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#7
Quote by philipisabeast
Okay i'm just doing a bit of maths revision and i'm slightly confused by the steps i need to go through to complete a quadratic equation

Q= Solve the following quadratic inequalities by first sketching appropriate graphs

a) (x+1) (x-4) < 0
= X[squared] - 3x - 4 < 0

So right off the bat you know that the X intercepts of the curve would be at -1 and 4

So thats listed, but how am i supposed to find the vertex? I've only seen how to do it with completed square equations


Start by drawing graph of y = (x + 1)(x - 4).

Crosses x-axis at -1 and 4.

To find y-intercept set x to zero, giving -4.

x value of turning point will be midway between -1 and 4, ie 3/2.

Now find y for that x:

y = x^2 - 3x - 4
=> (3/2)^2 - 3(3/2) - 4
=> 9/4 - 9/2 - 4
=> -(9/4) - 4
=> -(25/4)

Turning point is at (3/2,-25/4).

But turning point wouldn't be relevant to the question, if you just want to find for what values of x is (x +1 )(x - 4) < 0, ie between the x intercepts:

-1 < x < 4
Last edited by Malakian88 at Oct 11, 2007,