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  • Hate using Quadratic Formula ? This could help!

    Like it or not, we have to solve quadratic equations in our exams - mostly as part of another question, rather than a direct "find the roots" question. Unless we could factorize in one look, most of us go for the traditional quadratic formula which could take time, with its division and square root steps.

    Recently I came across another method - which was surprisingly simple, and was hidden in plain sight (at-least for me!)

    This method is pioneered by Prof. Po-Shen Loh (CMU)

    Here it goes.

    For the equation, ax^2 + bx + c = 0, we know sum of the roots = -b/a and product of roots = c/a.

    Let's take it one step ahead. consider m is the mid-point of the line connecting roots, we know it's (-b/a)/2.

    Let's explain further with a simple example

    Solve the roots for x^2 - 10x + 24 = 0

    I know, you have factorized it in your head, and know the roots already! But let's keep it aside, and try the new one.

    sum = 10
    mid point between the roots = 10/2 = 5
    which means the roots are at equal distance (say d) from 5 to either side.

    (Root1 = 5 - d) ----- d ------- m = 5 ----- d ------- (Root2 = 5 + d)

    so roots are (5 -d) and (5 + d)

    we know product of roots = 24,
    so (5 - d)(5 + d) = 24
    25 - d^2 = 24
    d^2 = 1

    d = 1 (as it is a distance, we take positive value)
    so roots are 5 -1 = 4 and 5 + 1 = 6

    next one.

    Solve the roots for 6x^2 - 19x + 10 = 0

    Maybe bit tricky to factorize for a poor mortal, and time consuming with the traditional formula.

    Will try with the above method

    sum of roots = 19/6
    mid point = 19/12
    roots are at same distance (say d) from 19/12
    so roots are (19/12 - d) and (19/12 + d).

    Product = 10/6
    (19/12 - d) * (19/12 + d) = 10/6
    (19/12)^2 - d^2 = 10/6
    d^2 = (19/12)^2 - 10/6 = (19^2 - 10 * 12 * 2)/144 = 121/144 = 11/12 (positive value only, as it's a distance)

    roots = (19/12 - 11/12) = 8/12 = 2/3 & (19/12 + 11/12) = 30/12 = 5/2

    one more ?

    solve the roots for 4x^2 + 3x - 1 = 0

    we know the drill :)

    sum = -3/4
    mid point = -3/8
    roots are equidistant (say d) from -3/8
    (-3/8 - d) and (-3/8 + d)

    product = -1/4
    (-3/8 - d) * (-3/8 + d) = -1/4
    (-3/8)^2 - d^2 = -1/4
    d^2 = 9/64 + 1/4 = (9 + 16)/64 = 25/64
    d = 5/8 (take only positive value, as it's a distance)

    roots = (-3/8 - 5/8) = -1, & (-3/8 + 5/8) = 1/4

    what if roots are imaginary?

    solve the roots for x^2 + x + 1 = 0

    sum = -1
    mid point = -1/2
    roots are equidistant (say d) from -1/2
    roots = (-1/2 - d) and (-1/2 + d)

    product = 1
    (-1/2)^2 - d^2 = 1
    d^2 = 1/4 - 1 = -3/4
    d = i√3/2
    roots are -1/2 ± i√3/2

    Can you solve the roots for x^2 - 2x + 8/9 = 0 (maybe, even without a pen! :) )

    Happy Learning!

    posted in Quant Primer
  • RE: Quant Boosters - Hemant Malhotra - Set 14

    @Yash-H-Chaudhari Could you please provide your solution?

    posted in Quant - Boosters
  • RE: Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved)

    @swati-jha Thanks Swati. Corrected the solution.
    Mistake was in distance values and is now clean, hopefully :)
    Happy learning!

    posted in Quant BBQ - Best of Best Questions
  • RE: CAT Question Bank - Games & Tournaments

    @mahisap I think the answer is:

    B
    D
    34

    What is your answer?

    posted in DILR - Booster
  • RE: Discussion Room : Quant

    @rohitchopra01 said in Discussion Room : Quant:

    If 76/(4+√7+√11)=p+q√7+r√11+s√77 then p+q+r+s=?

    Multiply both sides by (4 + 71/2 + 111/2)

    76 = (4 + 71/2 + 111/2)p + 71/2(4 + 71/2 + 111/2)q + 111/2(4 + 71/2 + 111/2)r + 771/2(4 + 71/2 + 111/2)s

    Expand out:
    76 = 4p + 71/2p + 111/2p + 71/24q + 7q + 771/2q + 111/24r + 771/2r + 11r + 771/24s + 111/27s + 71/211s

    Now get everything together:
    76 = (4p+7q+11r) + 71/2(p+4q+11s) + 111/2(p+4r+7s) + 771/2(q+r+4s)

    Rewrite the LHS:
    76 + 0 * 71/2 + 0 * 111/2 + 0 * 771/2 = (4p+7q+11r) + 71/2(p+4q+11s) + 111/2(p+4r+7s) + 771/2(q+r+4s)

    So we have the following system of equations:
    4p + 7q + 11r = 76
    p + 4q + 11s = 0
    p + 4r + 7s = 0
    q + r + 4s = 0

    4 equations in 4 unknowns. Solve.

    posted in Discussion Room
  • RE: Discussion Room : Quant

    @swanandk12 I think it is infinity. As x increases the value of F(x) will keep on increasing. So possible values of F(x) are infinite. Hence the sum of possible values F(x) is infinity. Thoughts?

    posted in Discussion Room
  • RE: Question Bank - 100 Algebra Questions From Previous CAT Papers (Solved)

    @visheshsahni

    No one is perfect here yaar and if you find a mistake in a solution and know the right answer, least thing to do is to share the solution. That's how a forum (should) function right ? :)
    Corrected the question.

    posted in Quant BBQ - Best of Best Questions
  • RE: Question Bank - 100 CAT level questions on Time, Speed and Distance topic

    t + t/12 = 9 (time taken in the return journey is 1/12th the time taken for the onward journey)
    t = 108/13 = 8.30 hours
    t + t/6 = 9.68 hours?

    posted in Quant BBQ - Best of Best Questions
  • RE: Question Bank - Algebra - Shashank Prabhu, CAT 100 Percentiler

    @shreyasnegi13
    There are 10 primes in the given range.
    4 also satisfies
    9 and 25 won't work because we would already have two 3s and 5s respectively in the numerator
    So 10 + 1 = 11 values.

    posted in Quant BBQ - Best of Best Questions
  • RE: Discussion Room : DILR

    @laavanya-ramaul

    120 = I + II + III (Where I - number of students who attended exactly one interview, II - exactly two interviews and III - exactly three interviews)
    I > II > III
    III is at-least 1

    a) maximum number of students who attended exactly 3 companies.
    We have to minimize I and II.
    Let say x people attended all three. and let x + 1 attended exactly two companies and x + 2 attended exactly one company.
    So 3x + 3 = 120
    x = 39

    posted in Discussion Room