With respect to the roots of x22x3=0, we can say that

A.

both of them are natural numbers

B.

both of them are integers

C.

one of the root is zero

D.

roots are not real

#### SOLUTION

Solution : B

Step 1:-   For x22x3=0, value of discriminant
D=(2)24(3)(1)=4+12=16.

Step 2:-  Since, D is a perfect square, roots are rational and unequal.

Step 3:- Solving the equation,
x=(2)+(16)2 or (2)(16)2
x=3,1
Thus, both of them are integers.

If the equation 9x2+6kx+4=0 has equal roots, then find the value of k =

A.

0

B.

2 , 0

C.

-2, 0

D.

± 2

#### SOLUTION

Solution : D

Given, the roots of quadratic equation 9x2+6kx+4=0 are real and equal.

Discriminant, Δ=b24acΔ=36k24(4)(9)=36k24(36)

The roots of quadratic equation are real and equal then Δ=036k24(36)=0k24=0k2=4k=4k=± 2

The equation is a quadratic equation : x+51=2x+10x6

A.

True

B.

False

#### SOLUTION

Solution : A

x+51=2x+10x6
(x + 5)(x - 6) = 2x + 10
x26x+5x30=2x+10
x23x40=0  is of the form of  ax2+bx+c

Taylor purchased a rectangular plot of area 634 m2. The length of the plot is 2 m more than thrice its breadth. The length and breadth respectively is _____ (approximate values).

A.

34.6 m & 11.20 m

B.

88 m & 24 m

C.

32 m & 16 m

D.

44.6 m & 14.20 m

#### SOLUTION

Solution : D

Let, Length = x

Given,
Area of Rectangle = 634 m2
Length : x
2 more that thrice : 3y+2
So, x=3y+2

634=xy634=(2+3y)y634=2y+3y23y2+2y634=0

This equation resembles the general form of quadratic equation ax2+bx+c=0.

Lets find the values of y satisfying the equation.(Roots of the equation)

y=b ±b2 4ac2a=2 ±22  4×3(634)2 × 3y=2 ±4 + 76086=2 ±76126y=2 ± 87.2466y=2+87.2466 or 287.2466y=14.20 or 14.87

Length is always positive.
y=14.20 m

x=2+3yx=2+3(14.20)=44.6 m

Length =44.6 m

Find the two consecutive positive integers, such that the sum of their squares is 365.

A.

17 , 18

B.

13 , 14

C.

9 , 10

D.

12 , 13

#### SOLUTION

Solution : B

Let the numbers be x &  x+1

Square of consecutive number: x2 & (x+1)2
Sum of square: x2+(x+1)2
x2+(x+1)2=365

Lets solve
x2+x2+2x+1=3652x2+2x364=0x2+x182=0
(Dividing by 2)

Factorise,
x2+14x13x182=0x(x+14)13(x+14)=0(x+14)(x13)=0x=13 & 14

Since numbers are positive integers, x=13

So, required consecutive numbers are 13 & 14.

Solve following equation for x:
x23x10=0

A.

-2 and -6

B.

10 and 5

C.

5 and -2

D.

7 and 6

#### SOLUTION

Solution : C

By using the method of factorisation,

x23x10=0
x25x+2x10=0

x(x5)+2(x5)=0

(x5)(x+2)=0

x=5 and x=2

Find the roots of the equation 5x26x2=0 by the method of completing the square.

A.

5±193

B.

3±195

C.

5

D.

3

#### SOLUTION

Solution : B

Multiply the equation by 5.

we get 25x230x10=0

(5x)2[2×(5x)×3]+323210=0
(5x3)2910=0(5x3)219=0(5x3)2=195x3=±19x=3±195

Roots of the equation are 3+195 & 3195

Find the numbers such that the square of number is 15 lesser than twice the number.

A.

8

B.

9

C.

-3

D.

5

#### SOLUTION

Solution : C and D

Let the number be x.
Square of number: x2
Lesser by 15: x215
of two times the number: x215=2x

We get, x215=2x
Lets solve, x22x15=0x25x+3x15=0x(x5)+3(x5)=0(x5)(x+3)=0x=5 & x=3

The number satisfying the given statement are 5 & 3.

If x=23 is a solution of the quadratic equation 7x2+mx3=0, find the value of m.

A.

m=16

B.

m=13

C.

m=16

D.

m=13

#### SOLUTION

Solution : C

Given: x=23 is a solution of the quadratic equation 7x2+mx3=0.

Substituting x=23 in the given quadratic equation we get

7(23)2+m(23)3=0

7(49)+m(23)3=0

289+2m33=0

28+6m27=0

m=16

For what value of m does the equation, x2+2x+m=0 have two distinct real roots?

A. m = 1
B. m < 1
C. m > 1
D. m = 2

#### SOLUTION

Solution : B

Step 1 : For, x2+2x+m=0,
a=1, b=2, c=m

We know that

D=b24ac
D=(2)24m
=44m

Step 2 : The roots of a quadratic equation are real and distinct only when D>0

Step 3 :  44m>0

4>4m

m<1