Write f(x) = 20x - 9 - 4x^2 in the form f(x) = c + (ax + b)squared

[richiesmasher]

f(x) = 20x-9-4x²

Derive f(x) in the form f(x)= c+(ax+b)²

I grouped like terms : -4x² +20x-9

I factored out -4 [x² +(20/-4)x -(5/2)² ]-9 +[(-4)-(5/2)² ]

Now I rewrote as a perfect square : -4[x-(5/2) ]² +16

Now I multiply the the bracket by -4 I get : (-4x+10)² +16

Rewrite as 16-(4x+10)²

But my book has : 16-(2x-5)² as the answer.

Where did I go wrong?

 

[Deleted member 4993]

f(x) = 20x-9-4x²

Derive f(x) in the form f(x)= c+(ax+b)²

I grouped like terms : -4x² +20x-9

I factored out -4 [x² +(20/-4)x -(5/2)² ]-9 +[(-4)-(5/2)² ]

Now I rewrote as a perfect square : -4[x-(5/2) ]² +16

Now I multiply the the bracket by -4 I get : (-4x+10)² +16

Rewrite as 16-(4x+10)²

But my book has : 16-(2x+5)² as the answer.

Where did I go wrong?

-4[x-(5/2) ]² +16

=-(2)² * [x-(5/2) ]² +16

= - (2*x - 2*5/2)² + 16 ← critical step

= -(2x -5)² + 16

= 16 - (2x -5)²

 

[chris84567]

Now I multiply the the bracket by -4 I get : (-4x+10)² +16

When you multiply the 4 you have to take the square root.

4(x+5/2)²
2*2(x+5/2)(x+5/2)(2x+5)(2x+5)
(2x+5)²

If don't then its like you're multiplying by 4² because you have 2 separate parts with the square.

 

[richiesmasher]

-4[x-(5/2) ]² +16

=-(2)² * [x-(5/2) ]² +16

= - (2*x - 2*5/2)² + 16 ← critical step

= -(2x -5)² + 16

= 16 - (2x -5)²

I see, but what happens to the square term in this term: -(2)² ?

I understand what happens if you just multiply by the 2 but how did you get rid of the square?

 

[tkhunny]

\(\displaystyle -4(x+1)^{2} = - (2^{2})(x+1)^{2} = -(2(x+1))^{2} = -(2x+2)^{2}\)

If there is a trick, here, it is that the negative sign does NOT come with the 4.