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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