Quadratic Identities (1 Viewer)

[aimzlouise]

Hey guys, I’m currently stuck on maths question so some help would be much appreciated.

3. If 2x²+x+1 and a9x+-1)(x-2) +b(x-1)+c are equal for all values of x, find a , b and c.
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Please show working. Thanks again people…

 

[tacogym27101990]

Hey guys, I’m currently stuck on maths question so some help would be much appreciated.

3. If 2x²+x+1 and a(x+-1)(x-2) +b(x-1)+c are equal for all values of x, find a , b and c.
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Please show working. Thanks again people…

just equate the coeffiecients
cant be bothered to actually do it
no maths for a while

 

[aimzlouise]

Thats why im stuck.... because i was reading the examples in the book and i just don't get it...

 

[Trebla]

Hey guys, I’m currently stuck on maths question so some help would be much appreciated.

3. If 2x²+x+1 and a(x+-1)(x-2) +b(x-1)+c are equal for all values of x, find a , b and c.
<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o> </o>
Please show working. Thanks again people…

Just expand the whole thing. Collect all x² terms and equate them to both sides. Do this similarly for x and the constant term. You end up with three simultaneous equations to solve for a, b and c.
This is just using the theorem that:
If ax² + bx + c = dx² + ex + f, then
a = d, b = e and c = f

 

[gurmies]

Hey, can you repost the question please, I can't really read what it says, i'll try to help

 

[Timothy.Siu]

Hey, can you repost the question please, I can't really read what it says, i'll try to help

If 2x²+x+1 and a(x+-1)(x-2) +b(x-1)+c are equal for all values of x, find a , b and c.

i guess i may as well do it then.

a(x^2-3x+2)+bx-b+c=ax^2-3ax+2a+bx-b+c=ax^2+x(b-3a)+(2a+b+c)

equation coefficients, a=2 b-3a=1 2a+b+c=1
sub a into the 2nd one, b-6=1 b=7 sub into third, 4+7+c=1 c=-10

i think..

 

[aimzlouise]

Hey thanks everyone for you're help