# 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.

= (a-1)^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.

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