Value of 'c' in quadratic equation.

[Johnc227]

The equation of a curve is y = ax² - 2bx + c, where a, b and c are constants with a being greater than 0.
(a) Find in terms of a, b and c the coordinates of the vertex of the curve.
(bi) Given that the vertex of the curve lies on the line y = x, find an expression for c in terms of a and b.
(bii) Show that in this case, whatever the value of b, c is equal or greater than -1/4a.

Part (a). I completed the square and found the vertex to be (b/a, c-b²/a)
Part (bi). When the vertex lies on y = x. Then b/a = c-b²/a. Rearranging for c = (b + b²)/a.

I am stuck with the final part (bii). I can see that after completing the square I end up with;
a(x-b/2a)² - b²/4a + c.

The constant part; - b²/4a + c looks to me an approach to -1/4a but I am drawing a blank so far.

This is a first post. I hope it is OK. And a grateful thanks to anyone looking at this.
John C

 

[Deleted member 4993]

The equation of a curve is y = ax² - 2bx + c, where a, b and c are constants with a being greater than 0.
(a) Find in terms of a, b and c the coordinates of the vertex of the curve.
(bi) Given that the vertex of the curve lies on the line y = x, find an expression for c in terms of a and b.
(bii) Show that in this case, whatever the value of b, c is equal or greater than -1/4a.

Part (a). I completed the square and found the vertex to be (b/a, c-b²/a)
Part (bi). When the vertex lies on y = x. Then b/a = c-b²/a. Rearranging for c = (b + b²)/a.

I am stuck with the final part (bii). I can see that after completing the square I end up with;
a(x-b/2a)² - b²/4a + c.

The constant part; - b²/4a + c looks to me an approach to -1/4a but I am drawing a blank so far.

This is a first post. I hope it is OK. And a grateful thanks to anyone looking at this.
John C

Treat the equation c = (b + b²)/a. as a quadratic in b.

What will be the condition for getting a "real" b from the equation above?

 

[Johnc227]

Dear Subhotosh Khan,
Thank you for spending the time to look at this problem to help me. I am very grateful.

I did treat the equation c = (b + b²)/a as a quadratic in b as you suggested;

b²/a +b/a - c = 0, I then completed the square for the quadratic in b which would give a minimum value for c.
It took me a day to realise that the coefficient was 1/a and not just a, (assuming that is correct of course).

1/a (b + 1/2)² -1/4 - c = 0 ;

(1/a) x (-1/4) = -1/4a


1/a (b + 1/2)² -1/4a - c = 0 indicating that c must be equal to or greater than -1/4a.

Again many thanks.
John C

 

[Deleted member 4993]

Dear Subhotosh Khan,
Thank you for spending the time to look at this problem to help me. I am very grateful.

I did treat the equation c = (b + b²)/a as a quadratic in b as you suggested;

b²/a +b/a - c = 0, I then completed the square for the quadratic in b which would give a minimum value for c.
It took me a day to realise that the coefficient was 1/a and not just a, (assuming that is correct of course).

1/a (b + 1/2)² -1/4 - c = 0 ;

(1/a) x (-1/4) = -1/4a


1/a (b + 1/2)² -1/4a - c = 0 indicating that c must be equal to or greater than -1/4a.

Again many thanks.
John C

Slightly different way:
c = (b + b²)/a

ac = b + b²

b² + b - ac = 0

Discriminant of the above quadratic equation must be ≥0 for real values of 'b'. So:

1^2 - 4*1*(-ac) ≥ 0

1 ≥ 4*1*(-ac)

Now multiply both sides by (-1/4) and flip the inequality condition

-1/4 ≤ ac

Divide both sides by 'a' (given a>0)

-1/(4a) ≤ c

or

c ≥ - 1/(4a)..... Those parentheses are very important