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Determining roots

2x+3=2x29|2x+3|=2x^2-9 how can I determine the roots?

Attachment not found
Graph looks like this
(edited 6 years ago)
Original post by joyoustele
2x+3=2x29|2x+3|=2x^2-9 how can I determine the roots?

Graph looks like this


Solve 2x+3=2x292x+3=2x^2-9 for x>32x > -\frac{3}{2} and (2x+3)=2x29-(2x+3)=2x^2-9 for x<32x < -\frac{3}{2}

Hint: you can factorise straight away
(edited 6 years ago)
Original post by joyoustele
2x+3=2x29|2x+3|=2x^2-9 how can I determine the roots?

Attachment not found
Graph looks like this


You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.
you need to find where the V shape crosses the U shape.
Reply 4
Original post by BuryMathsTutor
You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.


If I didnt have the graph, is there another way?
Reply 5
Original post by the bear
you need to find where the V shape crosses the U shape.


Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?
Reply 6
Original post by BuryMathsTutor
You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.


Yes, Thanks a lot
Reply 7
Original post by RDKGames
Solve 2x+3=2x292x+3=2x^2-9 for x>32x > -\frac{3}{2} and (2x+3)=2x29-(2x+3)=2x^2-9 for x<32x < -\frac{3}{2}

Hint: you can factorise straight away


Okay, TY :smile:
Original post by joyoustele
Okay, TY :smile:


The way to do it without the graph is to realise that 2x+3|2x+3| is actually just made up from two different straight lines, 2x+32x+3 and (2x+3)-(2x+3), where the changeover happens whenever 2x+3=02x+3=0 hence where the conditions on xx come from.

This means that, by definition, 2x+3={2x+3,x32(2x+3),x<32\displaystyle |2x+3| = \begin{cases} 2x+3, \qquad x \geq -\frac{3}{2} \\ -(2x+3), \quad x < -\frac{3}{2}\end{cases}.
(edited 6 years ago)
Original post by joyoustele
Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?


if the quadratic is inside a modulus then you can do that.
Reply 10
Original post by joyoustele
If I didnt have the graph, is there another way?

You can do these questions without a graph but you should always draw a graph in an exam - they’re usually quadratic/linear functions so easy to draw.

Doing it without a graph is risky unless you’re very confident with what you’re doing.
Original post by Notnek
You can do these questions without a graph but you should always draw a graph in an exam - they’re usually quadratic/linear functions so easy to draw.

Doing it without a graph is risky unless you’re very confident with what you’re doing.


Okay.
Thanks

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