2IIM Quant Notes  Linear & Quadratic Equations

rajesh_balasubramanian
Director, 2IIM Online CAT Preparation  IIT Madras  IIM Bangalore  CAT 100th percentile  CAT 2011, 2012 and 2014.
A quadratic function f(x) attains a maximum of 5 at x = 2. The value of function at x=0 is 1. Find the quadratic expression.
Any quadratic equation can be written in two ways, 1) C – a(xk)^2 and 2) a(xk)^2 – C.
These ways of writing the quadratic equation simply make use the fact that anything that is squared is always positive. So, for a minimum value of the function, we can express the quadratic function as sum of Squared Term + Minimum Value and for the Maximum value of the function, we can express it as Maximum Value – Squared Term.
So for this question, f(x) can be written as –> C – a(xk)^2.
Now, it is given that maximum value of f(x) is attained at x = 2. => C=5; since we will get the maximum value only if it is subtracted by the least quantity, k = 2.
5 – a(x2)^2. It is given that at x = 0, value of the function is 1. This means,
5 – a(2)^2 = 1 > 5 – 4a = 1 > a = 1.
So the expression is 5 – (x2)^2. So the quadratic expression is 5 – (x^2 – 4x + 4) => x2 + 4x + 1.
A system of equations has 3 equations
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.
If the system of equations has no solution, find k.This is a very interesting question. If we can generate one of the equations from the other two, we can then say that the system has infinite solutions. If we can generate one of the equations from the other two, but with the constant part alone being different, this would be akin to having parallel lines when we are dealing with 2 variables and that would result in the system having no solutions. So, lets look for that.
So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.
We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y ) = 2x + 3y . Then said a, b would give us k.Or, we are effectively solving for
3a + 4b = 2
4a + 5b = 3Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = 1. a = 2.
Now, k = 5a + 3b = 10 3 = 7.
If k = 7, this system of equations would result in no solutions.
If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + 7z = 6.
First equation * 2 – second equation would give us the equation 2x + 3y + 7z = 8.Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has no solution.
Yeah, the more mechanical, rather deceptively straightforwardlooking determinant method is also there. But where is the joy in that.
x^2 – 17x + p = 0 has integer solutions. How many values can p take?
To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the roots is p. Product of the roots is positive and so is the sum of the two roots.
So, both roots need to be positive.
So, we are effectively solving for
Number of positive integer solutions for a + b = 17.
We could have (1,16), (2, 15), (3, 14)……(8, 9) – There are 8 sets of pairs of roots. Each of these will yield a different product.
So, p can take 8 different values. Or, p can take 16 different values.
is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 – 17x = 0. This equation also has integer solutions.
So, p can also be zero.
So, number of possible values of p = 16 + 1 = 17.
Wonderful question – chiefly because there are two really good wrong answers one can get. 8 and 16. So, pay attention to detail. No point telling yourself “Just missed, I just didnt think of zero. I deserve this mark”. Being just wrong, will earn us a 1 instead of the honourable 0.
How many integer solutions exist for the equation x^2 – 8x – 48 =0?
One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly simpler method.
Note that x^2 is the same as x^2, so we can treat the equation as a quadratic in x.
Or, x^2 – 8x – 48 = 0
(x – 12) (x + 4) = 0
x could be 12. x cannot be 4.
If x could be 12, x can be 12 and 12.
Two possible solutions exist.3x + 4y = 33. How many integer values of (x, y) are possible?
A. 6
B. 3
C. 4
D. More than 6Let us rearrange the equation:
3x = 33 – 4y
Since x and y are integers, and since y is always positive regardless of the sign of y, this means that when you subtract a multiple of 4 from 33, you will get a multiple of 3.
Since 33 is already a multiple of 3, in order to obtain another multiple of 3, you will have to subtract a multiple of 3 from it. So, y has to be a positive or a negative multiple of 3.
y = 3, 3, 6, 6, 9, 9, 12, 12…etc.
For every value of y, x will have a corresponding integer value.
So there are infinite integer values possible for x and y.(x 3) (y + 4) = 12. How many pairs of integers (x, y) satisfy this equation?
4
10
6
8Product of two integers is 12. Further y + 4 > 4 . So, we know that one of the numbers has to be greater than 4.
So, we can have
1 x 12 => x = +4, y = +8
2 x 6 => x = +5, y = +2
3 x 4 => x = +6, y = 0
4 + 4 + 2 = 10 possible solutionsx + y = 8, x + y = 6.How many pairs of x, y satisfy these two equations?
A. 2
B. 4
C. 0
D. 1We start with the knowledge that the modulus of a number can never be negative, though the number itself may be negative.
The first equation is a pair of lines defined by the equations
y = 8 – x —— (i) {when y is positive}
y = x – 8 —— (ii) {when y is negative}
With the condition that x ≤ 8 (because if x becomes more than 8, y will be forced to be negative, which is not allowed)
The second equation is a pair of lines defined by the equations:
y = 6 – x —— (iii) {when x is positive}
y = 6 + x —— (iv) {when x is negative}
with the condition that y cannot be greater than 6, because if y > 6, x will have to be negative.
On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).
Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative
Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = 1. This satisfies both equations.
Only one solution is possible for this system of equations.What is the number of real solutions of the equation x^2  7x  18 = 0?
2
4
3
1Let us split this into two cases. Case 1, when x is greater than 0 and Case 2, when x is lesser than 0.
Case 1
x > 0. Now, x = x
x2 – 7x – 18 = 0
(x – 9) (x + 2) = 0
x is either –2 or +9.Case 2
x < 0. Now, x = –x
x^2 + 7x – 18 = 0
(x + 9) (x – 2) = 0
x is either –9 or +2.However, in accordance with the initial assumption that x < 0, x can only be –9 (cannot be +2).
Hence, this equation has two roots: –9 and +9.
Alternatively, we can treat this as a quadratic in x, the equation can be written as x^2 – 7 x – 18 = 0.
Or, (x – 9) (x + 2) = 0
x = 9 or –2. x cannot be –2.
x = 9, x = 9 or –9.Correct Answer: There are 2 solutions.
x^2  9x + k = 0 has real roots. How many integer values can 'k' take?
40
21
20
41Discriminant, D = 81 – 4k
If roots are real, D > 0
81 – 4k > 0
4k < 81
k < 20.25
Hence, –20.25 < k < 20.25
The integer values that k can take are –20, –19, –18 … 0 … 18, 19 and 20.
41 different values (Remember to include 0. Answer choice 'A' will definitely be 40)x^2  11x + p = 0 has integer roots. How many integer values can 'p' take?
6
4
8
More than 8To start with the discriminant should be a perfect square. Let the discriminant be 'D'.
From the quadratic formula: (−11± √D)/2, we see that the numerator has to be an even number for the roots to be integers.
This implies that the discriminant should be a perfect square and be square of an odd number. (Only then we will have odd + odd = even in the numerator)
D = 112 – 4p = 121 – 4p
4p cannot be negative => D can take values 121, 81, 49, 25, 9, 1
p can be 0, 10, 18, 24, 28, 30p can take 0, ±10, ± 18, ± 24, ± 28, ± 30
2x + 5y = 103. Find the number of pairs of positive integers x and y that satisfy this equation.
9
10
12
20Rearranging the equation, we get:
2x = 103 – 5yThis says that when you subtract a multiple of 5 from 103, you get an even number. You have to subtract an odd multiple of 5 from 103 in order to get an even number. There are 20 multiples of 5 till 100, ten of which are odd. (Note that you cannot subtract 105, or higher multiples as they result in a negative value for x.)
So, y can have ten integer values. x also has 10 integer values, each corresponding to a particular value of y.
y = 1, gives us a potential value for x, so do y = 3, 5, 7….19. y can take 10 values totally.
a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3.
Which of the following statements if true would imply that the above system of equations does not have a unique solution?
i. a1/a2 = b1/b2 = c1/c2 ≠ d1d2
ii. a1/a2 = a2/a3 ; b1/b2 = b2/b3
iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbersIf we have three independent equations, we will have a unique solution. In other words, we will not have unique solutions if:
The equations are inconsistent or
Two equations can be combined to give the third
Now, let us move to the statements.i. a1/a2 = b1/a2 = c1/c2 ≠ d1d2
This tells us that the first two equations cannot hold good at the same time.
Think about this:
x + y + z = 3;
2x + 2y + 2z = 5.
Either the first or the second can hold good. Both cannot hold good at the same time. So, this will definitely not have any solution.ii. a1/a2=a2/a3 and b1/b2=b2/b3
a1, a2, a3 are in GP, b1, b2, b3. This does not prevent the system from having a unique solution.
For instance, if we have
x + 9y + 5z = 11
2x + 3y – 6z = 17
4x + y – 3z = 15
This could very well have a unique solution.iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers
This gives us practically nothing. This system of equations can definitely have a unique solution.
So, only Statement i tells us that a unique solution is impossible.Equation x^2 + 5x – 7 = 0 has roots a and b. Equation 2x^2 + px + q = 0 has roots a + 1 and b + 1. Find p + q.
6
0
16
2Given, x^2 + 5x – 7 = 0 has roots a and b.
We know that, Sum of roots in a quadratic equation = a+b = (−5)/1 = 5.
Product of the roots = ab = (−7)/1 = 7.Now, The second equation 2x^2 + px + q = 0 has roots a + 1 and b + 1.
Sum of the roots = a+1+b+1 = a+b+2 = (−p)/2 = 5 + 2 = 3 = (−p)/2 => p = 6 => p =6.Product of the roots = (a+1)(b+1) = ab+a+b+1 = q/2. We know the values of ab and a+b. Substituting this, we get, 7+(5)+1 = q/2 => 11 = q/2 => q = 22.
Hence, p = 6 and q = 22. => p+q = 6 +(22) = 16.Sum of the roots of a quadratic equation is 5 less than the product of the roots. If one root is 1 more than the other root, find the product of the roots?
6 or 3
12 or 2
8 or 4
12 or 4Sum of the roots = (product of the roots) – 5
Let r1 and r2 be the two roots
r1=r2+1
Therefore,
r1+1+r2=(r1r2)∗r2−5
r2+1+r2=(r2+1)∗r2−5
2r2+6 = (r2)^2 + r2
(r2)^2−r2−6=0
Solving, we get r2 = 3 or 2
Therefore, r2 = 4 or 1
Hence product of the two roots is either 12 or 2.How many real solutions are there for the equation x^2 – 7x  30 = 0?
3
1
2
noneCase 1: x > 0
x^2 – 7x  30 = 0
( x  10 ) ( x  3 ) = 0
X = 10, x = 3
But x = 3 is not possible as we have considered x > 0, thus 1 solution for this case.
Case 2: x < 0
x^2 + 7x  30 = 0
( x + 10 )( x  3)=0
x = 10 and x = 3
Only x = 10 is permissible.
Thus this equation has 2 real solutionsAlternatively, we can think of the above as a quadratic in x
x^2 – 7x  30 = 0 can be factorized as
(x 10) ( x + 3) = 0
x cannot be 3, x can only be 10. X can take 2 real values.If (3x+2y22)2 + (4x5y+9)2 = 0 and 5x4y = 0. Find the value of x+y.
7
9
11
13Do not jump and start expanding the squares.
In equation (1), both the squares will yield either positive number or 0. Since sum of two positive number cannot be 0, each of the term must be 0
∴ 3x + 2y 22 = 0
Multiplying by 2, and adding both equation
6x + 4y  44 = 0
5x  4y = 0
=> x = 4
Putting the value of x in the equation
5 x 4 – 4y = 0
y = 5
∴ x+y = 9Let x^3 x^2 + bx + c = 0 has 3 real roots which are in A.P. which of the following could be true
b=2, c=2
b=1, c=1
b= 1, c = 1
b= 1, c= 1Given the roots are in A.P. so let ad, a, a+d be the roots
From equation, sum of roots = 1
Sum of two roots taken at a time = +b
Product of two roots = c
∴ (ad)+ (a)+ (a+d) = 1
=> 3a = 1
=> a = 1/3
Also, (ad)a+ a(a+d)+ (ad)(a+d) = b
=> a^2 – ad + a^2 + ad + a^2 – d^2 = b
=> 3a^2 – d^2 = b
=> 3 x 1/9 – d^2 = b
=> d^2 = b  1/3
Now, since d is a real number,
1/3 − b > 0 => b < 1/3Also,
(ad)(a)(a+d) = c
=> a(a^2 – d^2) = c
=> 1/3∗(1/9−d^2)=−c
=> c = d^2/3 − 1/27
=> c = (d^2 − 1)/27
=> d^2 = 3c + 1/9
Again, 3c + 1/9 > 0
c > −1/27 > 0∴ Only option (c) b= 1 , c =1 could be true .