CAT Previous Year Questions| Quadratic Equations

2x^2+kx+5=0
Since this Quadratic equation has no real roots, k^2<4(2)(5)
k^2<40
Since k is an integer, the possible values of k are (-6, -5, …, 5, 6)
x^2+(k−5)x+1=0
Since this Quadratic equation has real & distinct roots, (k−5)^2>4(1)(1)
(k−5)2>4
Since k – 5 is an integer, the possible values of k – 5 are (…, -4, -3, 3, 4, …)
the possible values of k are (…, 1, 2, 8, 9, …)
Putting both the inferences together, the possible values of k are (-6, -5, -4, -3, -2, -1, 0, 1, 2)
k can take 9 values.

Since the total donation amount is 15,250 rupees, there should be at least one 250 rupee note. The remaining 15000 can exist as thirty 500 rupee notes.

But the total number of notes will only be 31. We have 100 notes. So to keep the number of 500 notes as high as possible let’s convert the 500 rupee notes to 100 rupee notes. Every time we do this conversion we add 4 new notes.

From here, we can convert one 500 rupee note to two 250 rupee notes.

So, the maximum number of 500 rupee notes is 12

We are given two equations: 3x + 2|y| + y = 7 and x + |x| + 3y = 1
|x| or the modulus of x, is a function of x, that gives the magnitude of x.
|x| = -(x); if x is negative, and
|x| = x; if x is non-negative.
Therefore, depending on whether ‘x’ and ‘y’ are positive or negative, we assume the following cases for the two equations given.
Case (i): ‘x’ and ‘y’ are both positive.
|x| = x and |y| = y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x + 2y + y = 7
x + x + 3y = 1
3x + 3y = 7
2x + 3y = 1
Solving the two equations, we get x = 6 and y = -11/3.
Since this is contradictory to the assumption that y is positive, we discard this case.
Case (ii): ‘x’ and ‘y’ are both negative.
|x| = -x and |y| = -y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x – 2y + y = 7
x – x + 3y = 1
3x – y = 7
3y = 1
Solving the two equations, we get x = 22/9 and y = 1/3.
Since this is contradictory to the assumption that both x and y are both negative, we discard this case.
Case (iii): ‘x’ is negative and ‘y’ is positive.
|x| = -x and |y| = y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x + 2y + y = 7
x – x + 3y = 1
3x + 3y = 7
3y = 1
Solving the two equations, we get x = 2 and y = 1/3.
Since this is contradictory to the assumption that both x is negative, we discard this case.
Case (iv): ‘x’ is positive and ‘y’ is negative.
|x| = x and |y| = -y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x – 2y + y = 7
x + x + 3y = 1
3x – y = 7
2x + 3y = 1
Solving the two equations, we get x = 2 and y = -1.
This satisfies the assumption that x is positive and y is negative.
Hence x = 2 and y = -1
Therefore, x + 2y = 2 + 2(-1) = 0
Hence, x + 2y = 0.

For a quadratic equation, a x² + b x + c = 0.
the roots are given by x = −b±b2−4ac√2a
Since the roots, a, b, c are all rational and one of the roots, 2 + √ 3 is irrational, the other root of the equation will also be irrational and a conjugate of 2 + √ 3.
That is, the other root is, 2 – √ 3.
Sum of roots is given by −ba = (2 + √ 3) + (2 – √ 3) = 4.
And Product of roots is given by ca = (2 + √ 3) (2 – √ 3) = 2² – (√ 3)² = 4 – 3 = 1
But here the quadratic equation is a x² – b x + c = 0.
Therefore,
ba= 4 and ca = 1
b = 4a; c = a.
We are also given that b = c³
4a = a³
4 = a²
a = ∓2
|a| = 2

Two generic equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, will have a unique solution if and only if, a1/a2 ≠ b1/b2

.

We have the equations kx + y = 3 and 4x + ky = 4,

For them to have unique solutions, they must satisfy the condition

k/4 ≠ 1/k

k2 ≠ 4

k ≠ ±2

|k| ≠ 2

 The roots of a Quadratic Equation of the form, ax^2 + bx + c = 0

are given by x = −b±√[b^2−4ac]/2a

In order for these roots to be real, the portion under the square root should not be negative.

In other words ‘b^2 – 4ac’, determines wether the roots are real or imaginary.

Hence, rightly, it is called the Determinat(D).

If the Determinant, D is greater than or equal to 0, then the equation has real roots.

For D to be greater than or equal to 0, b^2 ≥ 4ac.

We are told that x^2 + mx + 2n = 0 and x^2 + 2nx + m = 0 have real roots,

That means, m^2 ≥ 4(2n) and (2n)^2 ≥ 4m.

Instead of trying to solve through equations, the two inequalities m2 ≥ 8n and n2 ≥ m.

Let’s try to solve inputting specific numbers.

If n = 1; m^2 ≥ 8; m ≥ 3

If m ≥ 3 at n = 1, n2 ≥ m stands invalid.

If n = 2; m^2 ≥ 16; m ≥ 4

If m ≥ 4 at n = 2, n2 ≥ m stands valid, if m = 4 and n = 2

Hence 4,2 is the smallest(and the only) pair that m,n can take.

So, the minimum sum of m and n is 4+2 = 6.

x = 2 and y = 19 (Not possible), Given the condition x ≥ y ≥ – 20
x should increase/decrease in the coefficient of y and same for y
x = 7 and y = 17 (Not possible)
x = 12 and y = 15 (Not possible)
x = 17 and y = 13 (Possible)
. y = 11
. y = 9
. .
. .
. .
. y = -19 (Can’t go any further)
By counting totally 17 values are there

We can have three possibilities for the equality of form xⁿ = 1
First n = 0, then x = 1 and then x = -1 and n = even (-1)^Even
Using first x = -1, and substituting in given equation
(-1)² -5(-1) + 7 = 8 (Acceptable)
Using second condition, x²
Solving x = 2 or x = 3 (Both are acceptable)
Last condition x2 -5x + 7 = -1
x² -5x + 8 = 0
No integer values,
So 3 values satisfy the equality

|x|² – 2|x| + |a – 2| = 0
|x|² – 2|x| + 1 = 0 is the square of a quadratic number
In the above equation the value of constant cannot be more than 1
So |a – 2| = 0 or = 1
|x|² – 2|x| = 0
|x|² = 2|x|
x = 0 or 2 or -2
For all these possibilities value of a = 2
|x|² – 2|x| + 1 = 0
(|x| – 1)² = 0
|x| = 1
So, x = 1 or x = -1
Then |a – 2| = 1, a = 3 or a = 1
Four combinations of (x,a) are possible already we have 3
Totally 7 pairs

(Aron) Px + sx + 2s = (Aditya) 2px + sx – 20 – 10x + 2s
Px = 10x + 20
x(p-10) = 20
Together they bought 3p pencils
Minimum value of p has to be 11
So, minimum pencils they bought together is 33

x + 1x = y
y² – 3y + 2 = 0
y ≥ 2 & y ≤ – 2
(y – 1) (y – 2) = 0
y = 1(Not possible) or y = 2
Now, x + 1x = 2
This has only one option x = 1. So, only one real root

(x²-7x+11)^{(x^{2}-13x+42)} = 1
Only possible thing is x^2-13x+42 = 0
Or x²-7x+11 = 1
Solving these two x²-13x+42 = 0
(x – 6) (x – 7) = 0
x = 6 or x = 7
x²-7x+10 = 0
(x – 2) (x – 5) = 0
x = 2 or x = 5
At this moment we have 4 values possible, But there is one more way we can arrive this
(-1)^Even = 1
So, x²-7x+11 = -1
(x – 3) (x – 4) = 0
x = 3 or x = 4
So, 4 + 2 = 6 values

can be rewritten as 
= 1 + 
1 + = 1 + 
n cannot be -3
has to be an integer
And this value has to be equal to 1, for n to be high
So,= 1
8 = n – 4
n = 12

Given quadratic equation is x² + bx + c = 0
Sum of roots = -b
-b = 4a + 3a
-b = 7a
Product of the roots = c
c = 4a x 3a
c = 12a²
b² = 49a² and c =12a²
b² + c = 49a² + 12a²
b² + c = 61 a²
Our answer must be a multiple of 61, which should be a perfect square
Going through the options,
549 = 61 x 9
549 = 61 x 3²
Thus, possible value of b² + c = 549

|x| (6x² + 1) = 5x²
We know that |x²| = x²
Let y = |x|
So, y² = x²
So, y (6y² + 1) = 5y²
6y² + 1 = 5y
6y² -5 + 1 = 0
y = or 
Since, y = |x|
y = or , or 
Since we have cancelled y in our first step, x = 0 is also a solution
So, Number of possible solutions = 4 + 1 = 5

Given that ∣x² – x – 6∣ = x + 2
Removing the modulus,
(x² – x – 6) = x + 2 (or) -(x² – x – 6) = x + 2
Solving (x² – x – 6) = x + 2
We get, x² – 2x – 8 = 0
(x-4) (x+2) = 0
x = 4 or -2
For the solution to be valid, the value of the equation must be positive
When x = 4,
x² – x – 6 = x +2
16 – 4 -6 = x + 2
6 – 2 = x
4 = x. Therefore, x = 4 works
For x = -2,
x² – x – 6 = x +2
4 + 2 – 6 = x + 2
0 – 2 = x
-2 = x. Therefore, x = – 2 works
Similarly, Solving -(x² – x – 6) = x + 2
-x² + x + 6 = x + 2
x² = 4
x = +2 or -2
For the solution to be valid, x² – x – 6 must be negative
When x = +2,
-x² + x + 6 = x +2
-4 + 2 + 6 = x + 2
+ 2 = x Therefore, x = 2 works.
Therefore, Product of distinct roots = 2 x -2 x 4 = -16