Higher Degree Equations & Waving Curve Method

Higher Degree Equations:

For higher degree equations of degree n.

We can use the following formulae:

Sum of roots taken one at a time (S₁) = - (coefficient of x^n-1/ coefficient of xⁿ)

Sum of roots taken two at a time (S₂) = coefficient of x^n-2/ coefficient of xⁿ

Sum of roots taken 3 at a time (S₃) = - (coefficient of x^n-3/ coefficient of xⁿ)

And so on with alternatively positive and negative sign.

Product of roots = (+ or -) constant term / coefficient of xⁿ

The positive sign is to be used for an even numbered degree equation and negative sign for an odd numbered degree equation.

Let’s take the below an example:

Example 1:

If p, q, r are the roots of the equation x³ + 6x² + 12x + 24 = 0, find the equation with the roots 2p, 2q and 2r.

We know, S₁ = p + q + r = - 6

• S₂ = 12
• P = - 24

Now, for the equation, whose roots are 2p, 2q and 2r.

S₁ = 2p + 2q + 2r = 2(p + q + r) = 2*(-6) = -12
S₂ = 2p*2q + 2q*2r + 2p*2r = 4*12 = 48
P = 2p*2q*2r = 8*(-24) = -192

Thus, required equation is x³ + 12x² + 48x + 192 = 0

Waving Curve Method:

This method is helpful in solving the inequalities involving higher degree.

Let’s take the below an examples:

Example 2:

Solve the inequality (x-3)(x-2)(x+1) ≤ 0

Step 1: First we find the zero’s of all factors (x-3)(x-2)(x+1). Zero’s can be found by putting all factors equal to zero.

In this case, the zero’s are 3,2 and -1, x-3 =0 we get x = 3;  x-2 = 0 we get x = 2; and x+1 = 0 we get x = -1

Step 2: Now mark the above said zero’s on a number line as shown:

Step 3: Now we will draw a curve starting from the greatest number (Extreme right hand side) in this case it is 3 and the curve would start from the upper side of the number line and should cross each of the marked point.

The regions covered by the curve above the number line are marked as +ve and the regions below the number line are marked as –ve.

Step 4: In our question we have to find the region ≤ so the answer is: (-∞, -1] U [2, 3].

This method is however, only applicable if the following conditions are met:

When the equality can be factorised in the form (x-a)(x-b)(x-c) where a, b anc are integers.
On the right hand side of the inequality, there should be a zero.
Negative sign should not be there with x. If it’s there, remove it by reversing the sign of the inequality.

Now, if the there is power of x in the numerator and the denominator has some factor(s), we will use the below example:

Example 3:

[(x-3)(x+2)(x+1)²/(x-2)] ≤ 0

First we mark the zero’s on the number line. Put a circle at a point that is derived from the factor which is in the denominator. Mark X on a point that is derived from a factor having even power. In this case it is -1.

Here, the curve will retrace its path- where X is marked. If it’s on the upper side, it’ll remain on upper side and similarly for lower side.

Now, the region is ≤ 0. We will not be including the points where circle is marked as it’s derived from the denominator. Thus the required region is: (-∞,-1] U (2,3]. 

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The chance of a particular event happening is called its probability. In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes.

In terms of formula it can be written as:

Probability = no. of Favorable outcomes / total no. of Possible outcomes.

The questions on probability can be on coins, dices, cards and arrangement of people.

It is important to understand a few terms before seeing some examples:

Tossing a Coin:

When a coin is tossed, there are two possible outcomes: Heads (H) or Tails (T). It can be concluded that the probability of the coin landing H is half or ½ and the probability of the coin landing T is also ½.

Throwing Dice:

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6.

The probability of getting an even numberon a single throw of dice is ½ as there are equal number of cases for an even number and an equal number of cases for an odd number.

Arranging People:

Let us say there are 3 people A, B and C. They are to be arranged in a row. There will be a total ofsix arrangements possible viz. ABC, ACB, BCA, BAC, CBA, CAB.

Now, if we have to compare only among A and B then out of the total arrangements half the number will have A coming before B i.e. ABC, ACB, CAB and in remaining half of the cases i.e. BAC, BCA and CBA B will be before A.

Hence, the probability of A coming before B is 50% i.e. ½ and that of B before A is also 50%.

Choosing a Card from a Deck:

There are 52 cards in a deck (not including Jokers). The number of black cards is 26 and the number of red cards is also 26. If a card is selected at random from the pack, then the probability of getting a red card is ½, as there is equal number of red and black cards.

Let us take a few examples to understand the concepts:

Example 1:

There are six marbles in a bag. Out of that, three are blue and three are red.

What is the probability that a blue marble is picked?

Solution:

Since there are equal number of blue and red marbles, so the chances of getting a red marble = ½ and also, the chances of getting a blue marble = ½.

Example 2:

Delegates of six different countries appear in a UNO meet for a conference with the UNO External Affairs Chief. Among them are one delegate each, from India and Pakistan. If each delegate meets the UNO chief one at a time.

What is the probability that India’s delegate meets the UNO’s Chief before Pakistan’s?

Solution:

As we have discussed earlier, out of the total 6! arrangements, in half the cases India’s delegate will be before Pakistan’s and in remaining half Pakistan’s will be before India’s. Hence, the required probability = ½.

Example 3:

There are three mothers with each mother having a child. They appear before the school selection committee for admission of their respective child. The committee decides to interview all six of them one by one.

What is the probability that every mother is interviewed before her child?

Solution:

Let us assume the names of the three mothers as M1, M2, M3, and their respective children as C1, C2 and C3. There will be total 6! arrangements possible.

Now, if we consider only M1 and C1 then in exactly half the number of cases, M1 is interviewed before C1 i.e. 720/2 = 360 cases, and in the rest 360 cases C1 will be interviewed before M1.

We reject this second case, hence, we are left with only 360 cases with M1 before C1. Again, among these 360 cases, half the number will have M2 before C2 i.e. 360/2 = 180 whereas the remaining 180 cases will have C2 before M2.

We reject this second case, hence, we are left with 180 cases having both the conditions satisfied i.e. M1 before C1 and M2 before C2. Again, among these remaining cases, half of them will have M3 before C3, and the other half will have C3 before M3.

So, we reject the second case again, and finally we are left with 180/2 = 90 cases, where M1 is before C1, M2 is before C2 and M3 is before C3 also.

Tricky Series asked in Various Exams

A series in which each term is formed by multiplying the corresponding terms of an A.P. and G.P. is called Arithmetico Geometric series. It is more popularly known as an A.G.P.The general or standard form of such a series is a, (a + d) r, (a + 2 d) r² and so on.

Sum of infinite number of terms of an A.G.P with |r| < 1 is

S ∞ = a / 1– r + rd / (1 – r)²

Where, a = first term
d = common difference of A.P.
r = common ratio of G.P.

Method of Differences:

In some series, the differences of successive terms (T_n and T_n-1) may be in some particular sequence- A.P., G.P., A.G.P. etc.

In these cases, the sum of series can be found by a technique known as "method of differences".

The sum of first n natural numbers-

The sum of squares of first n natural numbers-

The sum of cubes of the first n natural numbers-

Sum of first n odd natural numbers = 1+ 3 + 5 +……+ up to n terms = n².

Sum of first n even natural numbers = 2 + 4 + 6 +……+ up to n terms = n (n + 1).

For any series T_n = S_n – S_n–1

Example 1:

Evaluate 1 + 4/5 + 7/5² + 10/5³ +…… to infinite terms.

Solution:

Here, the series is an A.G.P, where 1, 4, 7, 10…….is an AP, with a = 1 and d = 3and 1, 1/5, 1/5²,….is a GP, with r = 1/5. For an A.G.P.,

S_∞ = a / 1– r + rd / (1 – r)², putting the values we get S_∞ = 35/16.

Example 2:

Find thesum to infinity of the series 1² + 2². x + 3². x² +..., |x| < 1.

Solution:

Though the given series is not an arithmetico-geometric series, however the differences (2² – 1²), (3² – 2²), ……form an AP. So, you can use theMethod of differences.

Let  S = 1 + 4x + 9x² + 16x³ +......∞. Multiply both sides with common ratio x of the GP.
Sx =  x + 4x² + 9x³ +........∞. Now, subtract the two equations
⇒ (1 – x) S = 1 + 3x + 5x² + 7x³ +.....∞ ……..(1)
Now, let R = 1 + 3x + 5x² + 7x³ +......∞, which is an arithmetico-geometric series with a = 1, d = 2 and r = x.
For an A.G.P., sum R = a / 1– r + rd / ( r– r)²
Substituting the values, you get R = 1+ x /(1 – x)²
Substitute R in (1), you get, (1 – x) S = 1+ x /(1 – x)²
⇒ S = 1+ x /(1 – x)³.

Example 3:

Find the sum of the series 2.5 + 5.8 + 8.11 +... upto n terms.

Solution:

This series is formed by multiplying the corresponding terms of sequences 2, 5, 8, ... and 5, 8, 11, ... both of which are A.P.s.

Now, nth term of first A.P. = 2 + (n – 1) 3 = 3n – 1.
and, nth term of second A.P. = 5 + (n – 1) 3 = 3n + 2.
Hence, nth term of the given series, T_n = (3n – 1) (3n + 2) = 9n² + 3n – 2.
Thus, the required sum, 
S_n = T_n = (9 n² + 3n – 2)
= 9  n² + 3 n – 2 = 9 n² + 3 n – 2nn.
⇒ 9 × n (n + 1)(2n + 1)/6 + 3 × n(n + 1)/2 – 2n.
On simplifying ⇒ n (3n² + 6n + 1).

Example 4:

Find the sum of the following series:

Solution:

There are 2007 terms in the given sequence and we want to evaluate S₂₀₀₇. Here, we will apply a shortcut called interpolation.

S₂ = T₁ + T₂ =

On interpolating, we can say S₃ = 4 – 1/4, S₄ = 5 – 1/5 ….so on.
Continuing the same logic S₂₀₀₇ = 2008 – 1/2008.