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Maximum and Minimum Value of a Quadratic Expression


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Quadratic equations constitute an important part of the IIT JEE Mathematics syllabus. A simple quadratic equation is of the form  

ax2+bx+c = 0, where x represents the variable or the unknown and a, b and c are the constants. The point to note here is that ‘a’ cannot be equal to zero as if a = 0, then the equation is no more a quadratic and will instead become linear.

What are the Maximum and Minimum Values of a Quadratic Expression?

We have already discussed the concept of quadratic expressions in the previous sections. It is a simple topic that can fetch you some direct questions which are scoring as well.

It has been observed that the graph of f(x) = ax2 + bx + c extends upwardly or downwardly in all cases accordingly to a > 0 or a < 0. Now, when graph extends upwardly, then the vertex determines the minimum of f(x) i.e. when a > 0, then at x = b/2a, f(x) attains its minimum equal to (–D/4a).

In the other case, when the graph extends downwardly, then the vertex determines the maximum of f(x), i.e., when a < 0, then at x = –b/2a, f(x) attains its maximum equal to (–D/4a). These same concepts are applicable in numerical which demand the computation of maximum or minimum values of a function.   

Some key Points

1. When a > 0, then maximum of f(x) does not exist and when a < 0, then minimum of f(x) does not exist. This happens because of R being the domain of the function f(x) = ax2 + bx + c.

2. The above graphically interpretation of maximum & minimum can be verified algebraically as under:

The given quadratic expression is

f(x) = ax2 + bx + c

We try to make the first term a perfect square and so we get the following equation

= [√ax + b/2√a]2 + (c-b2/4a)     (if a > 0)

The first term on R.H.S is a perfect square and so it can never be negative. It can at most be equal to zero for any x which happens at x = –b/2a.

If a > 0 then f(x)½min = -D/4a at x = -b/2a

If a < 0 then f(x) = a1x2 + bx + c where a = –a1

     = - (√(a1x) – b/2√a1)2 + (c + b2/4a1

The first term of R.H.S. can be at the most zero which is for x = –b/2a.

If a < 0 then f(x)½max = -D/4a at x = –b/2 

If the Quadratic is in the form

y = a(x-h)2 + k


In such a form of equation ‘k’ is the value at the vertex.

Here ‘k’ gives us the maximum or the minimum value of the function accordingly as ‘a’ is positive or negative.

Now, we discuss some of the illustrations based on these concepts:

View the following video for more on quadratic expressions 

Illustration: Find the greatest value of [(x+2)/ (2x2 + 3x + 6)] for real values of x?

Solution: The given function is [(x+2)/ (2x2 + 3x + 6)] 

Clearly, the function (2x2 + 3x + 6) is a quadratic function with a = 2 > 0 and so it will have its minimum at x = –b/2a = –3/4 and minimum value is

–D/4a = 9-48/ 4.2 = 39/8 

Moreover, D < 0 & 2x2 + 3x + 6 > 0 ∀ x

Hence, 0 < 1/ (2x2 + 3x + 6) < 8/39 ∀ x ∈ R

Hence, [(x+2)/ (2x2 + 3x + 6)] will have its maximum at x = –3/4

And the maximum value is given by (-3/4 + 2)/ (39/8) = 10/39

Illustration: Find the maximum or minimum value of x2 + x + 1.

Solution: Comparing the given equation with the general form ax2 + bx + c

We get, a = 1. Since the value of a > 0 so we will get a minimum value.

The minimum value is given by c-b2/4a = 1-12/4.1 = 3/4.

Illustration: Find the maximum or minimum value of -2(x-1)2 + 3.

Solution: As discussed above, this equation is of the form a(x-h)2 + k.

Here a = -2 and k = 3.

Now ‘a’ is negative which implies that the equation will have a maximum value. The maximum value in this case is given by k = 3.

Some Useful Results

1. If f(x) is a polynomial and a, b are the real numbers such that f (a). f (b) < 0 i.e., f(a) and f(b) are of opposite signs, then between a and b f(x) has at least one root and if it has more than one root between a and b then the no. of roots is odd. This theorem is basically used for obtaining approximate roots. 

2. If f (a). f (b) > 0 then between a and b there are either no roots or even no. of roots. An exception occurs when X axis turns out to be a tangent to the curve.

Lagrange’s Identity:  

The Lagrange Identity applies to two sets say {a1, a2, …. , an}and {b1, b2, …. , bn} of both real and complex numbers. In the general form it can be expressed as

The identity can also be written as  

(a12 + a22 + a32) (b12 + b22 + b32) =

(a1b1 + a2b2 + a3b3)2 + (a1b2 + a2b1)2 + (a2b3 + a3b2)2 + (a3b+ a1b3)2 

Illustration: If a, b, c, k are constants and a, b, g are variables subject to

a tan a + b tan b + c tan  g = k. then find the minimum value of tan2 a + tan2 b + tan2 g.

Solution: From the Lagrange’s Identity, we have,

(a2 + b2 + c2)(tan2 a + tan2 b + tan2 g)

= (a tan a + b tan b + c tan g)2 + (a tan b + b tan a)2 + (b tan g + c tan b)2 + (c tan a + a tan g)2

Also, (a2 + b2 + c2)(tan2a + tan2b + tan2g) ≥ (a tan a + b tan b + c tan g)2

or, tan2 a + tan2 b + tan2 g > k2/ (a2 + b2 + c2)              

Various Properties concerning ratios:

If a1/a2 = b1/b2 = c1/c2 then each of these ratios is equal to

1. (1a2 + mb1 + nc1)/ (1a2 + mb2 + nc2), here l, m, n ∈ R and l2 + m2 + n2 = 0

2. [(la1x + mb1x + nc1x)/ (la2x + mb2x + nc2x)]1/x , x ∈ R+

Conditions for Common Roots

1. Both the roots common:

Two quadratic common a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0   will have both the roots common if the coefficients of different powers is proportional

i.e., a1/a2 = b1/b2 = c1/c2

2. One root common:

Let a be the common root then

a2x2 + b1x + c1 = 0

a2x2 + b2x + c2 = 0

To eliminate a we have,

α2/ (b1c2 – b2c1) = α2/ (a2c1 – a1c2) = 1/ (a1b2 – b1a2

α2 = (b1c2 – b2c1) / (a1b2 – b1a2) , α2 = (a2c1 – a1c2)/(a1b2 – b1a2

This implies (b1c2 – b2c1) / (a1b2 – b1a2) = (a2c1 – a1c2)/(a1b2 – b1a2

Hence, (a2c1 – a2c2)2 = (a1b2–a2b1)(b1c2–b2c1)

This is the condition for the given two quadratic equations to have a common root.

Illustration: If the equation x2 + ax + b = 0 and x2 + bx + a = 0 have a common root, then the numerical value of a + b is …………

Solution: Let ‘a’ be the common root of the equation then it will satisfy the equation x2 + ax + b = 0.

This gives a2 + a a + b = 0

Subtracting we get,

a(a – b) – (a – b) = 0

(a – b)(a – 1) = 0

This gives a = 1, since (a-b) ≠ 0. 

Putting a = 1 in any of the above equations, we get,

a + b = –1.

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