Free Algebra 04 Practice Test - CAT
Question 1
Find the value of , given that
SOLUTION
Solution : C
Question 2
The value of a for which the quadratic equation 3x2+2(a2+1) x+ (a2-3a+2) =0 possesses real roots of the opposite sign lies in


SOLUTION
Solution : C
Answer = option (c)
To have 2 roots of the opposite sign, the product of the roots need to be negative and the equation should have real roots
4(a2+1)2 -12(a2-3a+2) ≥0 and (a2-3a+2)/3< 0 i.e. (a-1)(a-2)<0 or 1<a<2
Question 3
Given that P ≥ 1. What is the minimum value of P2 + 625/P2 ?
SOLUTION
Solution : C
Question 4
y = px2+12x+9 has real roots. What is the range for the value of p?
SOLUTION
Solution : C
Option (c)
Since the condition D≥0 is given (as the roots are real), we will find the value of p using this condition
D= b2-4ac≥0
122-(4xpx9) ≥0
144-36p≥0
144≥36p
p≤4
Question 5
Given that b3+mb2 +nb + c is divisible by (b-s) if s3 + ms2 + ns + c = 0Also given that d3+d2+ed+1 is divisible by (d-1) and d3-4d2+fd-3 is divisible by (d-3)What is the value of e+f ?
SOLUTION
Solution : C
Option (c)
d3+d2+ed+1 is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e
1+1+e+1=0 e=-3
Similarly put d=3 in d3-4d2+fd-3 to get the value of f
27-36+3f-3=0 3f=12 f=4
e+f= -3+4=1
Question 6
The roots of the equation (x+1)(x+8) +10 = 0 are given as p and q. Find the roots of (x+p)(x+q)-4=0
SOLUTION
Solution : C
Option c
(x+1)(x+8)+10 can be expanded as x2+9x+8+10=0 x2+9x+18=0
Sum of roots p+q= -9
Product of roots pq = 18
(x+p)(x+q)-4=0 can be written as x2+ (p+q)x+pq-4=0 x2-9x+18-4=0
x2-9x+14=0
Roots are 7 and 2
Question 7
Given that p and q are the roots of the equation x2 – ax +b =0 and Dn= pn+qn. Find the value of Dn+1
SOLUTION
Solution : A
Option (a)
Best way to proceed is by assumption of values.
Assume a quadratic equation. Let’s take x2+5x-6=0. Here the roots are p=1 and q=-6
Thus, a = sum of roots = -5 and b= product of roots = -6
D1= p1+q1 = -5
D0 = 2
D2 = 37
Assume n=1
We need to find Dn+1 = D2 = 37
Look in the answer options for 37
Option a is the only one which gives (-5)(-5)-(-6)(2) = 37
Question 8
if m>0, n>0 and p>0, then both the roots of the equation mx2+nx+p=0 are
SOLUTION
Solution : B
option (b)
x=(-b±√D)/2a
D=n2-4mp<n2 (as m>0 and p>0), the roots of this equation will be –ve. If D<0, then the roots will have negative real parts
Question 9
The equation has
SOLUTION
Solution : A
Option (a)
Squaring both sides
![]()
For x=5/4 LHS≠RHS
Hence x=5/4 is not a root of the equation.
Answer=a
Question 10
SOLUTION
Solution : D
Option (d)
Assume t= 2
The question changes to
[(5-√3i)(5+√3i)]/4
[25-(√3i)2]/4, since (a-b)(a+b)=a2-b2
(25+3)/4, since i2=-1
=7
Only answer option (d) gives 7 on substitution of t=2
Question 11
If a and b are the roots of the equation x2+px+1=0 and ; c and d are the roots of the equation x2+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?
SOLUTION
Solution : C
Shortcut: Assumption
Option (c)
Product of roots=1 is the only constraint.
Assume values that satisfy the constraint and substitute, take a= -1 and b= -1, then p= 2
Assume c= - 1/2 and d= - 2; q= 5/2
Question is = (a-c)(b-c)(a+d)(b+d)=9/4
Only option (c) satisfies this => q2 –p2 = 25/4 – 4 =9/4
Question 12
If the product of the roots for the equation x2 – 3kx +2e2ln(k) -1=0 is 7, then the roots of the equation are real for k= ?
SOLUTION
Solution : C
Answer = c
Product of roots = 2e2ln(k) -1 = 7; 2e2ln(k) = 8; e2ln(k) = 4; k2 = 4; k = ±2
For the roots to be real, k>0 k=2. to recheck CHECK for D>0; D= 9k2 -4 (2e2ln(k) -1) = 8>0
Question 13
There are 2 quadratic equations, a2 –a +p=0 and a2 – a + 3p=0 (p≠0). For what value of p will one root of the second equation = double the root of the first equation?
SOLUTION
Solution : C
Option (c)
Go from answer options
1) Take p=2
a2 –a +2=0 and a2 – a + 6=0
Roots are = 1+√(1+24)2 and √(1−24)2
Hence, this option is ruled out
Take p = -2
a2 –a -2=0 and a2 – a -6=0
Roots of 1st equation = 2,-1 and roots of second equation = 3, -2
Question 14
What is the value of X ?
SOLUTION
Solution : C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0
Question 15
Find the value of Z.
SOLUTION
Solution : C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0