Quadratic Equations CAT Previous Year Questions with Answer PDF

Since x & y are real numbers, (x + 2y) & (x − 2y − 1) are both real.

A square of a real number is always non-negative.

For this reason, for the equation: (x + 2y)² + (x - 2y - 1)² = 0 to be true, both (x + 2y) & (x - 2y - 1) must be equal to 0.

x − 2y − 1 = 0

x − 2y = 1

Sum of all possible values of x = 3 + (−5/2) = 1/2

“The price of a precious stone is directly proportional to the square of its weight.”

If W is the weight of the stone and P is the price of that stone, then P = k × W²
For the entire, unbroken stone, the price will be 18² × k = 324 k.

“If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.”

The minimum profit is achieved when the weights of the broken stones are close to each other, that is, the weights are 3, 4, 5, and 6 units.

In this case the combines worth of the four stones  = (3² + 4² + 5² + 6²) k = 86k

The maximum profit is achieved when the weights of the broken stones are far from each other, that is, the weights are 1, 2, 3, and 12 units.

In this case the combines worth of the four stones  = (1² + 2² + 3² + 12²) k = 158k

The difference in the total value = 2,88,000.

158k – 86k = 72k = 2,88,000

k = 4,000

So, the price of the original stone = 324 k = 12,96,000

We are given |x+a| + |x−1| = 2
|x - y| signifies the distance between the points x and y.
|x - (-a)| + |x−1| = 2
Is telling you that the sum of distances between the points x and (-a) and x and 1 is 2.
This implicitly means that the distance between (-a) and 1 can’t be more than 2 in the first place.
Imagine the distance between (-a) and 1 being more than 2.
For this explanation, let us assume that (-a) is to the left of 1.

In this case |x - (-a)| + |x−1| can never be equal to 2 and it will always be greater than 2.
So, the distance between (-a) and 1 can be exactly 2 units or less than 2 units.
Case i: distance between (-a) and 1 is less than 2 units.
Assume the distance between (-a) and 1 is less than 2 units and equal to some ‘x’ units
First of all x can’t be between (-a) and 1, because then |x - (-a)| + |x−1| will be lesser than 2.

x can be y units to the left of (-a) or y units to the right of 1 such that 2y + x = 2 units. But there will just be 2 such solutions and not infinite solutions.
Case ii: distance between (-a) and 1 can be exactly 2 units.
x can only be between (-a) and 1 and can not be anywhere else.
Because if x is not between (-a) or 1, |x - (-a)| + |x−1| will be greater than 2.

Since x can be anywhere between (-a) and 1, the value of |x - (-a)| + |x−1| will be exactly 2 units, this case satisfies the condition!
(-a) can be 2 units to the right of 1 or to the left of 1.
-a = -1
a = 1
(OR)
-a = 3
a = -3
The maximum value of a is 1.

Since a and b are a + b will also be a natural number, let’s call it k.
k * (k +1) = 42
The product of two consecutive natural numbers is 42.
k = 6.
a + b = 6

a = 2
b = 4.
2a + b = 4 + 4 = 8
2a + b = 8

Since r and (-r) are the roots of the cubic equation 5x³ + cx² − 10x + 9 = 0

10 r³ - 20 r = 0
r³ - 2r = 0
r (r² - 2) = 0

Assuming the third root of the cubic equation as k, we can reconstruct it as…

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

 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

For roots of any Quadratic equation to be real and distinct, D > 0
So, for x² − 4x - log₂A = 0,
D = (-4)² - (4 x 1 x (-log₂A)) > 0
16 + 4 log₂A > 0
4 + log₂A > 0
log₂A > -4
A >
A > (1/16)

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

Given expression, u² + (u − 2v − 1)² = −4v(u + v)
On expanding using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, we get
u² + u² + 4v² + 1 - 4uv - 2u + 4v = - 4vu - 4v²
2u² + 4v² - 2u + 4v + 4v² + 1 = 0
On rearranging and taking common we get,
2(u² - u) + 2 (4v² + 2v) = - 1
Add values on both sides to complete the squares inside parenthesis
2(u² – u + 1/4) + 2(4v² + 2v + 1/4) = -1 + 1/2 + 1/2
2(u - 1/2)² + 2(2v + 1/2)² = 0
Since they both are squares, they can’t be negative
So, u - 1/2 = 0 and 2v + 1/2 = 0
u = 1/2 and v = −1/4
U+3v = 

Let p and q be two roots of the equation
We have to find the minimum value of (p² + q²)
p² + q² = (p + q)² – 2pq ...(i)
Sum of the roots = (-b)/a.
Therefore, p + q = -(a + 3)
Product of the roots = c/a
Therefore, pq = -(a + 5)
Substituting these values back in (i)
p² + q² = [(-a + 3)² – 2(-a + 5)]
p² + q² = a² + 6a + 9 + 2a + 10
p² + q² = a² + 8a + 19
We have to find the minimum value of a² + 8a + 19.
By completion of squares, 19 can be split into 16 and 3
⇒ a² + 8a + 16 + 3
⇒ (a + 4)² + 3
The minimum possible value is 3 when a = -4