Quadratic Equations  Vikas Saini

2 basic formulae
for, equation ax^2 + bx + c = 0 ( a # 0 )
sum of roots =  b / a.
Product of roots = c / a.
Q. If equation Px^2 + Qx + R = 0, where P,Q,R integers and P+Q+R is equal to the smallest possible non negative integer. The roots are equal then find the sum of roots.
Solution :
Let's roots are m and m.
m + m =  Q / P .
2 m =  Q / P.
Q = 2Pm
m x m = R / P
m^2 = R / P .
R = Pm^2.
P + Q + R = 1.
P  2Pm + Pm^2 = 1.
P( 1  m)^2 = 1.
P = 1, m =2.
then Q = 4.
Sum of roots = 4.
Q. Let p and q be the roots of quadratic equation x^2  (a  2) x  (a+1) = 0.
find minimum possible value of p^2 + q^2.
Solution :
p + q = a  2.
pq =  (a + 1).
p^2 + q^2 = (p + q)^2  2pq
= (a  2)^2 + 2(a + 1)
= a^2  4a + 4 + 2a + 2
= a^2  2a + 6.
= (a1)^2 + 5.
if a = 1,
then p^2 + q^2 = 5.
Hence minimum value = 5.
Q. (a+b)x^2 + (c+d)x + e = 0 is a quadratic equation such that a:b = 1:3, b:c = 3:5, c:d = 5:11 and d:e = 11:7.
Find the absolute difference between the two roots of the equation.
Solution :
a = k, b = 3k, c = 5k, d = 11k, e = 7k.
(k+3k)x^2 + (5k+11k)x + 7k = 0.
4k x^2 + 16k x +7k = 0.
4x^2 + 16x + 7 = 0.
4x^2 + 14x + 2x + 7 = 0.
2x(2x+7)+(2x+7) = 0.
(2x+7) (2x+1) = 0.
x = 7/2 , 1/2.
absolute difference = 1/2 + 7/2 = 3.
Q. f(x) = px^2 + qx + r. If f(x) attains its maximum value at x = 2, then what is the sum of the roots of f(x) = 0.
Solution :
f(x) = px^2 + qx + r.
f '(x) = 2px + q.
put f '(x) = 0
x =  q / 2p.
function f(x) attains maximum value at x = q / 2p.
2 = q / 2p ... (i)
by putting f(x) = 0.
px^2 + qx + r = 0.
sum of roots = q / p.
from equation (i)
 q / 2p = 2.
 q / p = 4.
Q. The roots of the equation x^2 + bx  14 = 0 are m and n. If 4m+3n = 13, then which of the following must be true ?
(a) b = 31 / 12 (b) m = 2
(c) n = 14 (d) either a or b.
Solution :
m + n = b.
m x n = 14.
m = 14 / n.
4m + 3n = 13
4(14 / n) + 3n = 13.
= > 56 + 3n^2 = 13n.
= > 3n^2  13n  56 = 0.
= > 3n^2  21n + 8n  56 = 0.
= > 3n(n  7) + 8(n  7) = 0.
= > n = 7 , 8/3.
if n = 7, m = 2 and b = 5.
n = 8/3, m = 42 / 8 and b = 31/12.
Hence option d.
Q. x^2  35x + k = 0.
How many values of k exist such that both the roots of the equation are prime numbers ?
Solution :
suppose roots are m & n.
m + n = 35.
35 is an odd number and it is sum of one odd and one even number.
2 is only even number which is prime.
if m = 2, n =33.
but 33 is not prime here.
Hence there is no value of k for which both the roots of equation are prime numbers.
Q. Suppose k and 4 are two distinct roots of the quadratic equation ax^2 + bx + c = 0 with a > 0 and c > 0.
What you say about b ?
(a) b.
(b) b  4; b > 0 for other value of k
(c) b > 0 for all values of k
(d) b > 0 for k.
Solution :
two roots given 4 & k.
[(x  4 ) (x  k )] = ax^2 + bx + c.
= > [x^2  (4+k)x + 4k] = ax^2 + bx + c.
a > 0, c > 0.
suppose a = 1.
= > x^2  (k+4)x + 4k = x^2 + bx + c.
c = 4k.
b = (k+4)
(k+4) > 0, any value of k.
Hence option a.
Q. Suppose m & n are roots of equation x^2  10x +30 = 0.Then what is the value of (1+m+m^2) (1+n+n^2) ?
solution :
m + n = 10.
mn = 30.
(1+m+m^2) (1+n+n^2)
= (1+n+n^2)+(m+mn+mn^2)+(m^2+m^2n+(mn)^2)
= 1+(m+n)+(m^2 + n^2)+mn + (mn)^2 + mn (m+n)
= 1 + (m+n) + (m+n)^2  2mn + mn +(mn)^2 + mn(m+n)
= 1 + 10 + 10^2  2 x 30 + 30 + 30^2 + 30(10)
= 1281.
Q. If a & b are roots of x^2  kx + 72 = 0, and c & d are x^2  8x + k = 0.
a,b,c,d are natural numbers then
How many values can take a+b+c+d?
solution :
a + b = k.
ab = 72.
c + d = 8.
cd = k.
a+b = cd.
(a,b)= (1,72),(2,36),(3,24),(4,18 ),(6,12),(8,9).
(c,d )= (1,7),(2,6),(3,5),(4,4).
a+b = 73,38,27,22,18,17.
cd = 7,12,15,8.
no real value of a+b+c+d.
Q. If x = 1 / ( 4  2√3)
then find the value of x^4  4x^3 + 7x^2  6x + 7/4.
solution :
x = 1 / (4  3√3)
using rationalisation
x = 4 + 2√3 / 4
x = 1 + √3/2.
x^2 = 1 + 3/4 + √3 = 7/4 + √3
x^3 = 7/4 + √3 + 7√3/8 + 3/2
= 13/4 + √3 + 7√3 /8.
x^4 = 49/16 + 3 + 7√3/2.
x^4  4x^3 + 7x^2 6x + 7/4
= 97/3 + 7√3/2 13  4√3  7√3/2 + 49/4 + 7√3  6 3√3 + 7/4
= 17/16.
Q. What is the number of real solution of x^6 + 4x^3  12 ?
solution :
suppose x^3 = t & x^6 = t^2.
t^2 + 4t  12 = 0.
= > t^2 +6t 2t 12 = 0.
= > (t + 6) (t  2) = 0.
t = 6, 2.
x^3 = 6
x = 6, 6w , 6w^2.
x^3 = 2.
x = 2,2w,2w^2.
only 2 real solution.