Question 1

Roots of the equation x^2+b x-c=0(b, c 0) are

Accepted Answer

We know that, if roots are of same sign Product of root is + ve if roots are of opposite sign Product of roots is -ve Here, x^2+b x-c=0 (b, c 0 ) Product of roots = -c 1 =- ve Roots are of opposite sign.

Question 2

Rational roots of the equation 2 x^4+x^3-11 x^2+ x+2=0 are

Accepted Answer

Given equation can be reduced to a quadratic equation. 2 x^2+x-11+ 1 x + 2 x^2 =0 2 (x^2+ 1 x^2 )+ (x+ 1 x )-11=0 Put x+ 1 x =y 2 (y^2-2 )+y-11=0 2 y^2+y-15=0 y=-3 and 5 2 x+ 1 x =-3, x+ 1 x = 5 2 x^2+3 x+1=0,2 x^2-5 x+2=0 Only 2nd equation has rational roots as D =9 and roots are 1 2 and 2 .

Question 3

If 15^ and 30^ are the roots of the equation x^2+p x+q=0, then p q=

Accepted Answer

15^ = (45-30)^ = 3 -1 3 +1 15^ + 30^ =-p p= -4 3 ( 3 +1) q= 15^ 30^ = 3 -1 3 ( 3 +1) q= 3 -1 3 ( 3 +1) p q= -4 3 ( 3 +1) ( 3 -1) 3 ( 3 +1) p q= -4( 3 -1) 3( 3 +1)^2 = 10-6 3 3 p q= 10-6 3 3 Option (2) is correct.

Question 4

The roots of the given equation (p-q) x^2+(q-r) x+(r-p)=0 are :

Accepted Answer

Given equation is (p-q) x^2+(q-r) x+(r-p)=0 By using formula for finding the roots viz: -b b^2-4 a c 2 a , we get aligned & x= (r-q) (q-r)^2-4(r-p)(p-q) 2(p-q) \ & x= (r-q) (q+r-2 p) 2(p-q) = r-p p-q , 1 aligned

Question 5

The roots of the given equation (p-q) x^2+(q-r) x+(r-p)=0 are :

Accepted Answer

Given equation is (p-q) x^2+(q-r) x+(r-p)=0 By using formula for finding the root viz: -b b^2-4 a c 2 a , we get x= (r-q) (q-r)^2-4(r-p)(p-q) 2(p-q) x= (r-q) (q+r-2 p) 2(p-q) = r-p p-q , 1

Question 6

The range of the function f(x)= 3 x^2-4 x+5 is

Accepted Answer

f(x) is defined if 3 x^2-4 x+5 0 3 [x^2- 4 3 x+ 5 3 ] 0 3 [ (x- 2 3 )^2+ 11 9 ] 0 Which is true for all real x Domain of f(x)= (- , ) Let y= 3 x^2-4 x+5 y^2=3 x^2-4 x+5 i.e. 3 x^2-4 x+ (5-y^2 )=0 For x to be real , 16-12 (5-y^2 ) 0 y 11 3 Range of y= [ 11 3 , )

Question 7

Solution of 2^x+2^ |x| 2 2 is

Accepted Answer

2^x+2^ |x| 2 2 ...(i) Case I: x 0, then Eq. (i) becomes 2^x+2^x 2 2 2^x 2 x 1 2 Case II: x 2^x+2^ -x 2 2 t+ 1 t 2 2 , where 2^x=t t^2-2 2 t+1 0 x _2( 2 -1) Also, 0 The solution is . (- , _2( 2 -1) ) ] [ 1 2 , )

Question 8

If one root of the equation x^2+p x+12=0 is 4, while the equation x^2+p x+q=0. has equal roots, then the value of ' q ' is

Accepted Answer

4 is a root of x^2+p x+12=0 16+4 p+12=0 p=-7 Now, the equation x^2+p x+q=0 has equal roots. p^2-4 q=0 q= p^2 4 = 49 4

Question 9

If be a root of the equation 4 x^2+2 x-1=0, then the other root of the equation is

Accepted Answer

4 x^2+2 x-1=0 + =- 1 2 , =- 1 4 Also, 4 ^2+2 -1=0 as is a root and we have to prove that =4 ^3-3 . Now, 4 ^3-3 =4 ^2 -3 = (1-2 )-3 =-2 ^2-2 =- 1 2 [4 ^2+4 ]=- 1 2 [1-2 +4 ] =- 1 2 (1+2 )=- 1 2 - = Now, + =- 1 2 [from Eq. (i)] Hence, =4 ^3-3 .

Question 10

Let a, b be the solutions of x^2+p x+1=0 and c, d be the solution of x^2+q x+1=0. If (a-c)(b-c) and (a+d)(b+d) are the solution of x^2+a x+ =0, then is equal to

Accepted Answer

Since, a+b=-p, a b=1 .....(i) and c+d=-q, c d=1 Now (a-c)(b-c) and (a+d)(b+d) are the roots of x^2+a x+ =0 aligned & (a-c)(b-c)(a+d)(b+d)= \ & (a b-a c-b c+c^2 ) (a b+a d+b d+d^2 )= \ & \ 1-c(a+b)+c^2 \ \ 1+d(a+b)+d^2 \ = \ & (1+p c+c^2 ) (1-p d+d^2 )= \ & 1-p d+d^2+p c-p^2 c d+p c d^2+c^2-p c^2 d \ & +c^2 d^2= \ & 1-p d+d^2+p c-p^2+p d+c^2-p c+1= \ & [ c d=1] \ & 2+d^2+c^2-p^2= \ & aligned aligned 2 c d+c^2+d^2-p^2 & = \ (c+d)^2-p^2 & = \ q^2-p^2 & = [ (c+d)=-q] aligned

Question 11

Find the value of a such that the sum of the squares of the roots of the equation x^ 2 -(a-2) x-(a+1)=0 is least.

Accepted Answer

Let , be the roots of the equation. + =a-2 and =-(a+1) Now, ^ 2 + ^ 2 =( + )^ 2 -2 =(a-2)^ 2 +2(a+1)=(a-1)^ 2 +5 ^ 2 + ^ 2 will be minimum if (a-1)^ 2 =0, i.e. a=1.

Question 12

The number of real solutions of the equation (x-1)^ 2 +(x-2)^ 2 +(x-3)^ 2 =0 is

Accepted Answer

Equation holds if and only if x-1=0, x-2=0 and x-3=0 simultanesouly, which is not possible.

Quadratic Equation — Previous Year Questions & Solutions