### These CAT Quadratic Equations questions/problems with solutions provide you vital practice for the topic. The purpose of these posts is very simple: to help you learn through practice.

**Question 1:** If the equation *x ^{3} – ax^{2} + bx – a = *0 has three real roots then the following is true

**(a)**a = 1

**(b)**

*a*≠ 1

**(c)**

*b*= 1

**(d)**

*b*≠ l

### Answers and Explanations

The given equation *x*^{3 }– a*x*^{2 }+ b*x* – a = 0 can be rewritten as:

*x* (*x*^{2 }+ b) – a(*x*^{2} + 1) = 0

In case b = 1, then the equation becomes

(*x* – a)( *x*^{2} + 1) = 0

Now here x^{2} + 1 = 0 will give imaginary roots.

Hence, if the given equation has three real roots, then b ≠ 1.

**Question 2:*** m *is the smallest positive integer such that for an*y* integer *n ≥ m, *the quantit*y *n^{3} -7n^{2} +11n – 5 is positive. What is the value *of m *?

**(a)** 4

**(b)** 5

**(c)** 8

**(d)** None of these

### Answers and Explanations

**Answers: (d)**

Let *y* = *n*^{3} – 7*n*^{2} + 11*n* – 5

Now n = 1 is a root of the above equation.

Hence, it can be written as

*n*^{3} – 7*n*^{2} + 11*n* – 5 = (*n* – 1)(*n*^{2} – 6*n* + 5) = (*n* – 1)^{2}(*n* – 5)

Now, (*n* – 1)^{2} is alwa*y*s positive. So the whole expression will be positive if n – 5 > 0 i.e. n > 5.

Since n is an integer so its least value will be n = 6.

Now n ≥ m and m is also an integer.

Hence, the least value of m is 6.

**Question 3:** All the page numbers from a book are added, beginning at page 1. However, one-page number was mistakenly added twice. The sum obtained was 1000. Which page number was added twice?

**(a)** 44

**(b)** 45

**(c)** 10

**(d)** 12

### Answers and Explanations

** Answers: (c)**

Let the total pages be ‘n’. So we have {n(n+1)}/2 = 100

Since one page was added twice, so 1000 is not the actual sum but it is an increased sum.

We have n(n + 1) = 2000.

Now by hit and trial we can say that the value of n = 44 i.e. initially there were 44 pages and their sum was (44 x 45)/2 = 990

Since the given sum is 1000, so we can say that the page number 10 was added twice.

**Question 4:** Raman and Manoj attempted to solve a quadratic equation. Raman made a mistake in writing down the constant term. He ended up with the roots (4, 3). Manoj made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?

**(a)** (6, 1)

**(b)** (–3, –4)

**(c)** (4, 3)

**(d)** (–4, –3)

### Answers and Explanations

**Answers: (a)**

Since Raman made a mistake in constant term, so his product of roots will be wrong but his sum of roots will be correct.

Hence, the sum of roots is 4 + 3 = 7.

Manoj made a mistake in coefficient of ‘x’, so his sum of roots will be wrong but product of roots will be correct.

Hence, the product of roots is 3 × 2 = 6.

Hence, the required equation is x^{2} – 7x + 6 = 0.

The roots of this equations are 6 and 1.

**Question 5:** Let *p *and *q *be the roots of the quadratic equation *x ^{2}– (α – 2)x– *α

*–*1

*= 0.*What is the minimum possible value of

*p*?

^{2}+ q^{2}**(a)**0

**(b)**3

**(c)**4

**(d)**5

### Answers and Explanations

**Answers: (d)**

We have the equation x^{2}– (α – 2)x– α –1 = 0. its roots are p and q.

So, we have sum of roots = *p*+*q* = a–2 and the product of roots, pq = –a –1

Now p^{2}+q^{2} = (p+q)^{2} – 2pq = (a–2)^{2} + 2(a+1)= a^{2} +4 – 4a + 2a + 2 = (a – 1)^{2} + 5

Since (α – 1)^{2} is a perfect square, so its minimum value will be 0, when α = 1.

In that case, the minimum value of p^{2} + q^{2}will be 5.

Great