# Worked-out Problems

Example-1
If the roots of the quadratic equation x2 â€“ 4x â€“ log3N = 0 are real, then find the minimum value of N?
Solution
Since the roots of the given equation are real, so D â‰¥ 0
Or, 16 + 4 log3N â‰¥ 0, or, log3N â‰¥ âˆ’ 4
Or, N â‰¥ 3-4
So, the least value of N = 1/81

Example-2
Find the real values of N for which the quadratic equation 2x2 â€“ (N3 + 8N â€“1) x + N2 â€“ 4N = 0 will have the roots of an opposite sign.
Solution
The roots of the given equation will be of opposite signs only if they are real and the product of the roots is negative.
Or, D â‰¥ 0, and the product of the roots < 0
Or, (N3 +8Nâ€“1)2â€“8 (N2â€“ 4N) â‰¥ 0 and
Or, N2 â€“ 4N < 0. Hence 0 < N < 4

Example-3
If p and q are the roots of the quadratic equation 2x2 + 6x + N = 0 (N < 0), then what is the maximum value of
Solution
p + q = âˆ’3, and pq = N/2
Now, N < 0, hence D = 36 â€“ 4N > 0. So, p and q are real.
Again,
So, the maximum value of

Example-4
For which value of k does the following pair of equators yield a unique solution of x such that the solution is positive? (CAT 2005)
x2 âˆ’ y2 = 0
(x âˆ’ k)2 + y2 = 1
1. 2
2. 0
Solution
(3) y2 = x2
2x2 â€“ 2kx + k2 â€“ 1 = 0
D = 0
â†’ 4k2 = 8k2 â€“ 8
â†’ 4k2 = 8
â†’ k =

Example-5
Let
1. 3
2.
Solution
(c)

Now, put the values from the options.
Only the 3rd option satisfies the condition.
Alternatively, the value of x will be more than 2, which is given only in option 1 and option 3. Since it is only slightly more than 3, option 2 will be the answer.

Example-6
In a quadratic equation ax2 + bx + c = 0, ab and c are positive. Graph of this equation cuts X-axis at two distinct points. If both the roots of this equation are real, what can be said about the signs of both the roots?
Solution
Method 1
Let us assume p and q are the roots of the equation.

Product of the roots = p q = c/a = + ve[Both c and a are positive]

Hence, p and q are both positive, or both are negative.

Sum of the roots = p + q = âˆ’ b/a = âˆ’ ve

Hence, p and q are both negative.

Method 2
Since the roots are real and distinct, D > 0

Obviously p is negative.

In this case, we need to compare b and
so q will also be negative.

Method 3
Using Descartesâ€™ Theorem, sign of f(x) = + + +, so there will be no positive roots and x = 0 is not a solution of this equation.

So, it can be concluded that both the roots will be negative.