Question Bank  Algebra  Hemant Malhotra


@hemant_malhotra 1, x =0

@hemant_malhotra 1, x = 0?

@hemant_malhotra @zabeer b) 8 ?

@hemant_malhotra approach for this ?

@hemant_malhotra Ans 2. values 0, 1 ?

@hemant_malhotra can you please post the approach?

@hemant_malhotra 24 ?

@hemant_malhotra how can we do it with graphs ?

@hemant_malhotra c = 8 at 1/15

@hemant_malhotra 3 + 3 + 5 = 11

@hemant_malhotra 17?

@hemant_malhotra 110?

@hemant_malhotra approach

@hemant_malhotra a=b=c=4

@DhruvAgarwal correct!
@VikrantGarg @HarikrishnaShenoy Will try to explain this concept. A very useful and important one.
Here, Number of critical points = 1 + 2 + 3 + ... + 20 = 20 x 21/2 = 210
Minimum value of the expression is the Median which is at x = 1/15Need more explanation? Read on!
x  1 = x  1 ( 1 term)
2x  1 = x  1/2 + x  1/2 (2 terms)
3x  1 = x  1/3 + x 1/3 + x  1/3 (3 terms)
and so on..
20x  1 = x  1/20 + x  1/20 + x  1/20 + ... (20 terms)So x  1 + 2 x  1 + 3 x  1 + ... + 20 x  1 = x  1 + x  1/2 + x  1/2 + x  1/3 + x 1/3 + x  1/3 + ... + x  1/20 + x  1/20 + x  1/20
Why we are doing this ? so that every terms is in the form of x  a
How many terms are there ?
1 + 2 + 3 + 4 + ... + 20 = 210Now learn this concept
Minimum value of x –a + x – b + x – c... will be at value of x for median = 0
Here, Minimum value is at median, and as we have even numbers, it would be at 106th term
as we know sum of first 14 numbers is 105, 106th term would be of the form x  1/15 and the value of x = 1/15.To ensure you got this concept, let's solve some more.
Find the minimum value of x  2 + x  1 + x + 11
It is already in ascending order.
n is odd. Minimum will occur at x = middle value = 1
minimum value = 1 + 12 = 13Find the minimum value of x  1 + x  5 + x  25 + x  125 + x  625
In order. and minimum occurs at x = middle value = 25
minimum value = 24 + 20 + 0 + 100 + 600 = 744Find the minimum value of x  1 + 2x  1 + 3x  1 + .... + 100x  1
Similar to our original problem and minimum occurs at x = 1/71. Solve for yourselfClear with this concept ?

As we know a, b and c are the roots of the given equation, we know abc, ab + bc + ca and a + b + c right away!
For ax^3 + bx^2 + cx + d = 0
Sum of the roots =  b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = d/aFor our equation, 3x^3 + 2x^2 + x + 3 = 0
a + b + c = 1 * 1/3 = 2/3
ab + bc + ca = 1/3
abc = 3/3 = 1So we need to express the given expression in a form which constitutes the known forms.
(1/a^2) + (1/b^2) + (1/c^2) = a^2b^2 + a^2c^2 + b^2c^2/(abc)^2  (1)
(ab + bc + ac)^2 = a^2b^2 + a^2c^2 + b^2c^2 + 2ab^2c + 2a^2bc + 2abc^2
= a^2b^2 + a^2c^2 + b^2c^2 + 2abc (a + b + c)
a^2b^2 + a^2c^2 + b^2c^2 = (ab + bc + ac)^2  2abc(a + b + c)  (2)sub (2) in (1)
(1/a^2) + (1/b^2) + (1/c^2) =
[(ab + bc + ac)^2  2abc(a + b + c)]/(abc)^2
= [1/9  4/3]/(1)^2
= 1/9  4/3 = 11/9

As it is mentioned as unique positive integers.. (thanks @AmarRajput)
xyz = 45
45 = 1 * 3 * 3 * 5
(x, y, z) could be (3, 3, 5), (3, 15, 1), (1, 9, 5), (1, 1, 45) (unordered pairs)
In this set, least sum for unique integers is 15 for (1, 9, 5)
(any other thoughts?)

@hemant_malhotra
f(1) < 0
f(0) < 0
f(1) > 0
f(2) > 0
So one real solution between 0 and 1

We will split the equation in a way that we get only xyz terms in RHS.
x^2 + 4xy + 4y^2 + 2z^2
multiply all terms = x^2 * 4xy * 4y^2 * 2z^2 = (xyz)^2 * 32xy.
This can't be resolved as we don't know xySo we need to rearrange the terms so that it yields the RHS in terms of xyz (which we know). We know there is x^2, y^2 and z^2 available so we dont want to disturb that balance. We have extra xy, so we will split 2z^2 as z^2 + z^2 which will give some z without disturbing the balance of square terms. As we have 2 z's now outside, we need 2 x and 2 y too. So we will split 4xy as 2xy + 2xy. So the final equation becomes
(x^2 + 2xy + 2xy + 4y^2 + z^2 + z^2)
Now let's apply AM ≥ GM
(x^2 + 2xy + 2xy + 4y^2 + z^2 + z^2)/6 ≥ (16 * x^4 * y^4 * z^4)^1/6
≥ 6 * (16 * (32)^4))^1/6
≥ 6 * (2^24)^1/6
≥ 6 * 16
≥ 96
minimum value is 96