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Solved Examples on Quadratic Equations

Illustration 1: The set of all real numbers x for which x2 - |x + 2| + x > 0 is (2002)

1. (-∞, -2) ∪ (2, ∞)                                          2. (-∞, -√2) ∪ (√2, ∞)

3. (-∞, -1) ∪ (1, ∞)                                          4. (√2, ∞)

Solution: The condition given in the question is x2 - |x + 2| + x > 0

Two cases are possible:

Case 1: When (x+2) ≥ 0.

Therefore, x2 - x - 2 + x > 0

Hence, x2 – 2 > 0

So, either x < - √2 or x > √2.

Hence, x ∈ [-2, -√2) ∪ (√2, ∞)     ……. (1)

Case 2: When (x+2) < 0

Then x2 + x + 2 + x > 0

So, x2 + 2x + 2 > 0

This gives (x+1)2 + 1 > 0 and this is true for every x

Hence, x ≤ -2 or x ∈ (-∞, -2)       ...…… (2)

From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

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Illustration 2: If a, b and c are the sides of a triangle ABC such that

x2 – 2(a + b + c)x + 3μ (ab + bc + ca) = 0 has real roots, then (2006)

1. μ < 4/3                                                             2. μ > 5/3

3. μ ∈ (4/3, 5/3)                                                 4. μ ∈ (1/3, 5/3)

Solution: It is given in the question that the roots are real and hence, D ≥ 0

This gives, 4(a+b+c)2 - 12μ(ab+bc+ca) ≥ 0

Hence, (a+b+c)2 ≥ 3μ (ab+bc+ca)

So, a2+b2+c2 ≥ (ab+bc+ca)( 3μ-2)

Hence, 3μ-2 ≤ (a2+b2+c2)/(ab+bc+ca)

Also, cos A = b2 + c2 - a2 /2bc 

And so, cos A < 1

This means, b2 + c2 - a2 < 2bc

Similarly, c2 - a2 - b2 < 2ca

And a2 + b2 - c2 < 2ab

So, a2 + b2 + c2 < 2(ab + bc + ca)

Therefore, (a2+b2+c2)/(ab+bc+ca) < 2

Hence, using the obtained equations, we get

3μ -2 < 2

So, this gives μ < 4/3.

Illustration 3: If x2 – 10ax -11b = 0 has ‘c’ and ‘d’ as its roots and the equation x2 – 10cx -11d = 0 has its roots a and b , then find the value of a+b+c+d. (2006)

Solution: We have two equations x2 – 10ax -11b = 0

And x2 – 10ax -11b = 0

Now using the concepts of sum of roots we obtain the following relations,

a+b = 10c and c+d = 10a

So, we get (a-c) + (b-d) = 10(c-a)

This gives, (b-d) = 11(c-a) ….. (1)

Since, ‘c’ is a root of x2 – 10ax -11b = 0

Hence, c2 – 10ac -11b = 0 …..(2)

Similarly, ‘a’ is a root of the equation x2 – 10cx -11d = 0

So, a2 – 10ca -11d = 0    …… (3)

Now, subtracting equation (3) from (2), we get

(c2 - a2) = 11(b-d)           …….. (4)

Therefore, (c+a) (c-a) = 11.11(c-a)   …. (From eq(1))

Hence, this implies c+a = 121

Therefore, a+b+c+d = 10c + 10a

                                 = 10(c+a)

                                 = 1210.

Hence, the required value of (a+b+c+d) = 1210.


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