Thread: quadratic equation

Please help me with this one.

Show that x^2 + 3x + 5 > 0, for all values of x.

and

Show that 2x^2 + 6 > 4x + 1, for all values of x.

How do i show it,????

Originally Posted by mazy22a

Please help me with this one.

Show that x^2 + 3x + 5 > 0, for all values of x.

and

Show that 2x^2 + 6 > 4x + 1, for all values of x.

How do i show it,????

(1) Try completing the square .
(2) Rearrange it and try completing the square also .

Originally Posted by mazy22a

Please help me with this one.

Show that x^2 + 3x + 5 > 0, for all values of x.

Mr F says: y = x^2 + 3x + 5 is a positive parabola and has no x-intercepts (I leave the proof of no x-intercepts for you). Therefore ....

and

Show that 2x^2 + 6 > 4x + 1, for all values of x.

Mr F says: 2x^2 + 6 > 4x + 1 => 2x^2 - 4x + 5 > 0. So you need to show that 2x^2 - 4x + 5 > 0. The approach is the same as for the question above.

How do i show it,????

..

Originally Posted by mr fantastic

..

So basically i draw a graph.

And the graph will show that there is no x intercept.

Okay, cheers.

Originally Posted by mazy22a

So basically i draw a graph.

And the graph will show that there is no x intercept.

Okay, cheers.

How do you propose to draw the graph? Not the lazy way - using a graphics calculator - I hope.

To draw the graph by hand (which is what you should be attempting to do) you need to calculate the x-intercepts. Use the quadratic formula to solve x^2 + 3x + 5 = 0. What do you notice about the discriminant ....? Therefore .....

Originally Posted by mr fantastic

How do you propose to draw the graph? Not the lazy way - using a graphics calculator - I hope.

To draw the graph by hand (which is what you should be attempting to do) you need to calculate the x-intercepts. Use the quadratic formula to solve x^2 + 3x + 5 = 0. What do you notice about the discriminant ....? Therefore .....

the discirminant is negative, and less than 0.

D = -11