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3 weeks ago

How do I show that ax²+bx+c=0 can be written in the quadratic formula

The answer says (x+b/2a)²-(b/2a)²+c/a=0

But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead

The answer says (x+b/2a)²-(b/2a)²+c/a=0

But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead

Original post by Bird1234

How do I show that ax²+bx+c=0 can be written in the quadratic formula

The answer says (x+b/2a)²-(b/2a)²+c/a=0

But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead

The answer says (x+b/2a)²-(b/2a)²+c/a=0

But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead

Dividing through by a:

x^2 + b/a x + c/a = 0

Then when you complete the square for e.g. x^2+4x+5 you halve the b coefficient to give (x+2)^2 - 4 + 5

So in this case you halve b/a = b/2a to give (x+(b/2a))^2 - (b/2a)^2 + c/a

Original post by Notnek

Dividing through by a:

x^2 + b/a x + c/a = 0

Then when you complete the square for e.g. x^2+4x+5 you halve the b coefficient to give (x+2)^2 - 4 + 5

So in this case you halve b/a = b/2a to give (x+(b/2a))^2 - (b/2a)^2 + c/a

x^2 + b/a x + c/a = 0

Then when you complete the square for e.g. x^2+4x+5 you halve the b coefficient to give (x+2)^2 - 4 + 5

So in this case you halve b/a = b/2a to give (x+(b/2a))^2 - (b/2a)^2 + c/a

Ohh OK thank you

Reply 3

3 weeks ago

Alternatively, reverse engineer by expanding the result also works...

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