Problem Solving

What is the minimum value of
f(x) = -5 + (x + 7)^2, and at what value
of x does it occur?

I solved above question with below approach.
- The minimum value for f(x) will be 0 and for that we would need (X+7)^2 = 5.
- So, I thought solving the equation x^2 + 14x + (49-5)=0 will give me the required values of x.

Apparently, I am wrong and the right answers are -7, -5.

Can you explain where have I gone wrong?

Thanking you!

baalok88 wrote: - The minimum value for f(x) will be 0

Why does the minimum value of f(x) have to equal 0 ? Can't it be less than 0 ?

Looking at (x+7)^2 you can see it can only be positive so you want to keep the result as low as possible in order to minimize f(x). Can it = 0?

What if you made x equal to -7. That would make it 0. Then f(x) would be -5.

Your mistake was starting with an unexamined assumption that f(x) cannot be less than 0

We can write any quadratic as

ax² + bx + c

If a > 0, the quadratic opens up. So the minimum will come when the two roots are the same. (If you visualize the quadratic, you get the minimum where there isn't anything to mirror the point at which you're at.)

Since the roots are of the form

x = (-b ± √(b² - 4ac))/2a

They'll be equal when

x = (-b + √(b² - 4ac))/2a = (-b - √(b² - 4ac))/2a

or when x = -b/2a.

Since your quadratic is

x² + 14x - 44 = 0

The minimum comes when x = -14/2 = -7. Then, plugging in x = -7, we get -5 + (-7 + 7)², or -5.

baalok88 wrote: I solved above question with below approach.
- The minimum value for f(x) will be 0 and for that we would need (X+7)^2 = 5.
- So, I thought solving the equation x^2 + 14x + (49-5)=0 will give me the required values of x.

You're on the right track, but you're thinking of a quadratic that already equals 0. This is an easy mistake to make, since the first introduction to quadratics in algebra (and for many people, the ONLY quadratics they worked with!) are when quadratic = 0.

But it isn't a law that the quadratic = 0: it could be more or less than that.

As for the second part, you were solving for the two ROOTS of the quadratic, another really common chore in high school algebra. Those are helpful, but they don't necessarily give you the minimums, only the point (if any) where the parabola (which is the name for the visualization of the quadratic) crosses the x-axis. (There are also parabolas that cross the y-axis in two places, but those don't tend to show up on the GMAT, so we won't bother with them.)

For much more on this, check out Purple Math.