# Free Basic Algebra - 02 Practice Test - CAT

The roots of the equation a2x2+abx=b2,a0 are:

A. Imaginary
B. Real
C. Zero
D. Can’t say

#### SOLUTION

Solution : B

D=(ab)2+4a2b2=5a2b2

a2b2 has to be greater than or equal to 0 (i.e.,) D>0. So,roots are always real.

Find the value of k so that sum of the squares of roots of equation x28x+k=0  is 30.

A. 12
B. 14
C. 16
D. 17

#### SOLUTION

Solution : D

Let the roots be a,b.

ab=k,a+b=8

a2+b2=30

(a+b)2=64,a2+b2+2ab=64

30+2k=64,k=17

Let p, q be the roots of the equation x23x+2=0. The quadratic equation having roots as p+2 and q+2  is:

A. x2x=0
B. x27x+12=0
C. x27x12=0
D. x2+7x+12=0

#### SOLUTION

Solution : B

Substitute x by (x-2)
The equation becomes: (x2)23(x2)+2=0,x27x+12=0.

Let p, q be the roots of the equation x23x+2=0.

The quadratic equation having roots 2p and 2q is:

A. x23x+4=0
B. x26x8=0
C. x26x+4=0
D. x26x+8=0

#### SOLUTION

Solution : D

Substitute x by( x2)

The equation becomes: (x2)23(x2)+2=0, x26x+8=0.

Let p, q be the roots of the equation x23x+2=0.

The quadratic equation having roots 1p and 1q  is:

A. 2x23x+1=0
B. 2x23x1=0
C. 2x23x+2=0
D. 2x2+3x+1=0

#### SOLUTION

Solution : A

Soln:

Interchange ‘a’ and ‘c’ in the equation ax2+bx+c=0

The equation becomes: 2x23x+1=0.  Hence option (a)

If p, q are roots of an equation x2+5x+2=0 , then find out the value of pq+qp.

A. 152
B. 252
C. 212
D. 17

#### SOLUTION

Solution : C

Sum of the roots: p+q=5

Product of the roots: pq=2

pq+qp=(p2+q2)pq=[(p+q)22pq]pq=[(p+q)2]pq2=2522=212

If the roots of the equation 3x2+bx+3=0 are in the ratio 4:3 , then find out the value of b.

A. 732
B. 122
C. 73
D. 4

#### SOLUTION

Solution : A

For a quadratic equation ax2+bx+c=0 , if the roots are in ratio m:n then mnb2=(m+n)2ac.

In the equation given:

4×3×b2=(4+3)2(3×3)

12b2=49×9

b=7×3(23)=732

If equations x2kx21=0 and x23kx+35=0;k>0 have a common root, then k is equal to:

A. -4
B. 5
C. 4
D. -4 or 4

#### SOLUTION

Solution : C

The equations x2kx21=0 and x23kx+35=0 have a common root.

So, equating the two equations:

x2kx21=x23kx+35

2kx=56

k=28x

Putting in the equation:

x22821=0

x2=49

x=+7

or k=+4,As,k>0,k=4.

If the roots of the equation: ax2+bx+c=0,a>0  be each greater than unity, then:

A. a+b+c>0
B. abc>0
C. a+b+c>0
D. a+b+c=0

#### SOLUTION

Solution : A

Both the roots are greater than 1 and a > 0. So, the graph of the equation will be: f(x)=ax2+bx+c

f(1)>a+b+c>0

Find out the value of k for which the expression x2+(k+5)xk5=0 has two distinct real roots.

A. (,9][5,)
B. (,9)(5,)
C. (,9)(3,)
D. (-9, -5)

#### SOLUTION

Solution : B

For the equation given
D=(k+5)2+4(1)(k+5)=k2+10k+25+4k+20=k2+14k+45

For the equation to have two distinct roots:
D>0

k2+14k+45>0 (k+9)(k+5)>0 Value of k will be (,9)(5,)

The minimum value of the expression;  a+1a;a>0 is:

A. -2
B. 2
C. 0
D. 1
E. Can’t be determined

#### SOLUTION

Solution : B

Soln:

a+1a=(a1a)2+2

For minimum value of a+1a ,we want minimum value of (a1a)2 that will be zero.

So, minimum value of a+1a will be 2. Hence option (b)

2nd method:- By,  A.M  G.M   a+1a2(a×1a thus a+1a>2 (i.e.,)(a+1a)min will be 2 .  option (b).

If p, q are the roots of the equation x23x+2=0 , find the equation which has roots as (2p + 1) and (2q + 1).

A. x28x+12=0
B. 2x28x+15=0
C. 2x27x+14=0
D. x28x+15=0

#### SOLUTION

Solution : D

Given equation: x23x+2=0
with roots p, q

When the roots become 2p and 2q,
the equation becomes:
(x2)23(x2)+2=0,x26x+8=0

When the roots become
(2p+1) & (2q+1)
the equation becomes:

(x1)26(x1)+8=0,x28x+15=0.
Hence option (D)

For the equation x8+6x75x43x2+2x5=0, determine the maximum number of real roots possible.

A. 4
B. 3
C. 8
D. 7

#### SOLUTION

Solution : A

f(x)=x8+6x75x43x2+2x5

Check the number of positive roots= no. of sign changes in f(x) = 3

Check the number of negative roots = no. of sign changes in f(-x)

f(x)=x86x75x43x22x5 . No. of sign changes = 1

Now check for zero as a root. f(0)=50

So, maximum number of real roots = 3 + 1 = 4. Hence option (a)

Suppose a and b are two roots of the equation x2(α4)x+α=0. Find out the maximum possible value of 5aba2b2.

A. 0
B. 1614
C. 39
D. 5

#### SOLUTION

Solution : B

a and b are the roots of the equation

a+b=(α4),ab=α

5aba2b2

5ab(a2+b2)

5ab[(a+b)22ab]

5α[(α4)22α]

5α[α210α+16]

α2+15α16=(α2+15α16)=(α2+16αα16)=[α(α+16)1(α+16)]=(α+16)(α1)

So,the maximum value =D4a=[1524(1)(16)]4×1=1614.

A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchange their places. The number is ___.

#### SOLUTION

Solution :

Let the two digit number be ab.

a×b=12

When 9 is added to the number

10a+b+9=10b+a

9a9b+9=0

ab+1=0

a12a+1=0

a2+a12=0

(a+4)(a3)=0

a = -4 is not accepted. So, a = 3, b = 4

The number is 34 .