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

Attachment 730072 Graph looks like this
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RDKGames
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(Original post by joyoustele)
|2x+3|=2x^2-9 how can I determine the roots?

Graph looks like this
Solve 2x+3=2x^2-9 for x > -\frac{3}{2} and -(2x+3)=2x^2-9 for x < -\frac{3}{2}

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

Attachment 730072 Graph looks like this
You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.
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the bear
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you need to find where the V shape crosses the U shape.
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joyoustele
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(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?
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joyoustele
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(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?
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joyoustele
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(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
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joyoustele
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(Original post by RDKGames)
Solve 2x+3=2x^2-9 for x > -\frac{3}{2} and -(2x+3)=2x^2-9 for x < -\frac{3}{2}

Hint: you can factorise straight away
Okay, TY
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RDKGames
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(Original post by joyoustele)
Okay, TY
The way to do it without the graph is to realise that |2x+3| is actually just made up from two different straight lines, 2x+3 and -(2x+3), where the changeover happens whenever 2x+3=0 hence where the conditions on x come from.

This means that, by definition, \displaystyle |2x+3| = \begin{cases} 2x+3, \qquad x \geq -\frac{3}{2} \\ -(2x+3), \quad x < -\frac{3}{2}\end{cases}.
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the bear
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(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.
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Sir Cumference
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(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.
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joyoustele
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(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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