# Algebra - Quadratic Equations - Previous Year CAT/MBA Questions

The best way to prepare for Algebra - Quadratic Equations is by going through the previous year **Algebra - Quadratic Equations cat questions**.
Here we bring you all previous year Algebra - Quadratic Equations cat questions along with detailed solutions.

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**IIFT 2019 QA | Algebra - Quadratic Equations cat Question**

Consider the equation: |x – 5|^{2} + 5|x – 5| – 24 = 0. The sum of all the real roots of the above equation is:

- (a)
2

- (b)
3

- (c)
8

- (d)
10

Answer: Option D

**Explanation** :

Let |x − 5| = k.

Hence, the equation can be rewritten as: k^{2} + 5k − 24 = 0

Solving the equation, we get k = 3, −8.

But, |x − 5| cannot be negative, so k ≠ −8.

∴ |x − 5| = 3

So, x − 5 = ±3.

∴ x = 8 and x = 2

Required sum = 8 + 2 = 10.

Hence, option (d).

Workspace:

**XAT 2018 QADI | Algebra - Quadratic Equations cat Question**

Two different quadratic equations have a common root. Let the three unique roots of the two equations be A, B and C - all of them are positive integers. If (A + B + C) = 41 and the product of the roots of one of the equations is 35, which of the following options is definitely correct?

- (a)
The common root is 29.

- (b)
The smallest among the roots is 1.

- (c)
One of the roots is 5.

- (d)
Product of the roots of the other equation is 5.

- (e)
All of the above are possible, but none are definitely correct.

Answer: Option C

**Explanation** :

A, B and C are positive integers.

Let the common root is A.

∴ A × B = 35

This is possible when (A, B) is (1, 35) or (35, 1) or (5, 7) or (7, 5)

C = 41 – A - B

For each of these 4 possibilities value of C = 5 or 5 or 29 or 29

In each of these 4 possibilities, one of the roots is definitely 5.

Hence, option (c).

Workspace:

**IIFT 2018 QA | Algebra - Quadratic Equations cat Question**

The roots of quadratic equation *y*^{2 }– 8*y* + 14 = 0 are α and β. Find the value of (1 + α + β^{2})( 1 + β + α^{2})

- (a)
419

- (b)
431

- (c)
485 + 3√22

- (d)
453 + √22

Answer: Option B

**Explanation** :

The roots of the equation y^{2} – 8y + 14 = 0 are α and β.

∴ Sum of roots = α + β = −(−8)/1 = 8 and Product of roots = αβ = (14)/1 = 14

∴ (1 + α + β^{2}) (1 + β + α^{2}) = 1 + α + β^{2} + β + αβ + β^{3} + α^{2} + α^{3} + α^{2}β^{2}

= 1 + (α + β) + (αβ) + (αβ)^{2} + (α^{2} + β^{2}) + (α^{3} + β^{3})

= 1 + (α + β) + (αβ) + (αβ)^{2} + (α + β)^{2} – (2αβ) + (α + β)^{3} – (3αβ)(α + β)

= 1 + 8 + 14 + (14)2 + (8)^{2} – 2(14) + (8)^{3 }– 3(14)(8)

= 23 + 196 + 64 – 28 + 512 – 336 = 431

Hence, option (b).

Workspace:

**XAT 2017 QADI | Algebra - Quadratic Equations cat Question**

If x and y are real numbers, the least possible value of the expression 4(x - 2)^{2} + 4(y - 3)^{2} – 2(x - 3)^{2 }is:

- (a)
-8

- (b)
-4

- (c)
-2

- (d)
0

- (e)
2

Answer: Option B

**Explanation** :

4(x – 2)^{2} + 4(y – 3)^{2} – 2(x – 3)^{2}

= 4(x^{2} + 4 – 4x) + 4(y – 3)^{2} – 2(x^{2} + 9 – 6x)

= 4(y – 3)^{2} + 2x^{2} – 4x – 2

Now, (y – 3)^{2} least value would be 0.

The least value of 2x^{2} – 4x – 2 would be at x = -(-4)/(2×2) = 1

The least value of 2x^{2} – 4x – 2 = 2(1)^{2} – 4(1) – 2 = -4

The least value of the expression would be = – 4.

Hence, option (b).

Workspace:

**XAT 2015 QA | Algebra - Quadratic Equations cat Question**

Devanand’s house is 50 km West of Pradeep’s house. On Sunday morning, at 10 a.m., they leave their respective houses.

Under which of the following scenarios, the minimum distance between the two would be 40 km?

**Scenario I:** Devanand walks East at a constant speed of 3 km per hour and Pradeep walks South at a constant speed of 4 km per hour.

**Scenario II:** Devanand walks South at a constant speed of 3 km per hour and Pradeep walks East at a constant speed of 4 km per hour.

**Scenario III:** Devanand walks West at a constant speed of 4 km per hour and Pradeep walks East at a constant speed of 3 km per hour.

- (a)
Scenario I only

- (b)
Scenario II only

- (c)
Scenario III only

- (d)
Scenario I and II

- (e)
None of the above

Answer: Option A

**Explanation** :

Scenario 1: Devanand walks East at a constant speed of 3 km per hour and Pradeep towards South at a constant speed of 4 km per hour,

Let the two walk for x hours.

Distance travelled by Devanand and Pradeep is 3x km and 4x km respectively.

Thus, we have

∴ (AP)^{2} + (CP)^{2} = (50 – 3*x*)^{2} + (4*x*)^{2} = 402

Solving this, we get *x* = 6

Thus, after 6 hours, the minimum distance between the two would be 40 km.

The distance between the two will increase in other two scenarios.

Hence, option (a).

Workspace:

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