Examples of Common Substitutions or Transformations

Algebraic Manipulation (Part 3) : Substitutions (Part 1)

The link to the last article is here.

Here's a simple example :

This doesn't seem difficult. However, a square root is in the way. What should we do?

Let's make it disappear! Let $y = \sqrt{x^2 + 18x + 45}$. Immedaitely, the equation becomes $y^2 - 15 = 2y$ or $(y - 5)(y + 3) = 0$. But square roots can't be negative! So, $y = 5$. Substituting back gives $x^2 + 18x + 20 = 0$, which by Vieta's Formula have product of roots $20$.

Again this problem is not hard. After the substitution, the problem is instantly solved. Substitution is a common and useful technique to get rid of nasty radicals. It is also very useful in inequalities, as we'll see in Substitutions (Part 2).

Algebraic Manipulation (Part 4) : Substitutions (Part 2)

The last article is Algebraic Manipulation (Part 3) : Substitutions (Part 1)

In this article we look at a tougher problem that requires substitution.

The following solution will not be complete, it will only show the usage of Substitutions in this problem.

The fractions are ugly, so wishful thinking led us to substitute $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$ to transform the original inequality to

$\frac{x^2}{y + z} + \frac{y^2}{x + z} + \frac{z^2}{x + y} \ge \frac{3}{2}$, with the help of the fact that $xyz = 1$. This might be trivial for those who knows Titu's Lemma. Indeed, a direct application of T2's Lemma gives $\frac{x^2}{y + z} + \frac{y^2}{x + z} + \frac{z^2}{x + y} \ge \frac{(x + y + z)^2}{2(x + y + z)} = \frac{x + y + z}{2} \ge \frac{3}{2}$, where the last inequality follows from AM-GM and $xyz = 1$.

Note that Titu's Lemma states that $\frac{a_1^2}{x_1} + \frac{a_2^2}{x_2} + ... + \frac{a_n^2}{x_n} \ge \frac{(a_1 + a_2 + ... + a_n)^2}{x_1 + x_2 + ... + x_n}$, where $a_1, a_2, ..., a_n$ are real numbers, and $x_1, x_2, ..., x_n$ are positive real numbers.

The next post will be on difference of squares.

Problem Solving: Substitution

Many times when factoring polynomials, or solving complicated algebra equations, it is useful to make a clever substitution to make life easier for us. Example

Solve $x^4+5x^2+6=0$.

This is a quartic. Normally, we try to find two binomials that multiply to this. With some guessing and checking, we find that $(x^2+3)(x^2+2)=0$ so our roots are $\pm 3i$ and $\pm 2i$. However, there is a simpler way to do this. Let $a=x^2$. We can express the equation in terms of $a$. $$x^4+5x^2+6=\left(x^2\right)^2+5(x^2)+6=a^2+5a+6.$$ Now this is a quadratic we know how to handle. We can factor this as $(a+2)(a+3)=0$ so $a=-2,-3$. Solving for $x$ yields $$\begin{align} x^2=-2\implies x&=\pm 2i \\ \\ x^2=-3\implies x&= \pm 3i \end{align}$$ This is a much easier way to do this. Of course this is just the beginning of how substitution can greatly simplify problems.

The solution to the previous problem is as follows. Let $q=\sqrt{4x-3}$. Now all we have to solve is $$q+\frac{10}{q}=7$$ which is simple. Multiplying by $q$ for each side yields $$\begin{align}q^2+10=7q \\ &\implies q^2-7q+10=0 \\ &\implies (q-5)(q-2)=0 \\ &\implies q=2,5.\end{align}$$

Now we equate each value of $q$ to $\sqrt{4x-3}$. So $$\begin{align}&q=2\implies 2=\sqrt{4x-3}\implies 4=4x-3\implies x=\frac{7}{4} \\ \\ &q=5\implies 5=\sqrt{4x-3}\implies 25=4x-3\implies x=7.\end{align}$$ Therefore, $\boxed{x=\frac{7}{4},~7}$.

Solution to previous problem

Let $u=\log x$. We have $$\begin{align} 12u^2+11u+2=0 &\implies 12u^2+3u+8u+2=0 \\ &\implies 3u(4u+1)+2(4u+1)=0 \\ &\implies (4u+1)(3u+2)=0 \\ &\implies u=-\frac{1}{4},~-\frac{2}{3} \\ \\ &\implies \begin{cases} \log x=-\frac{1}{4} \\ \log x=-\frac{2}{3} \end{cases} \\ \\ &\implies \begin{cases} x=10^{-1/4} \\ x=10^{-2/3} \end{cases} \\ \\ &\implies \boxed{ \begin{cases} x\approx 0.56 \\ x\approx 0.22 \end{cases}} \end{align}$$

The topic of inequalities holds its own importance. Here in these series of notes, I would try to unravel those.: $$Geometric \ Inequalities$$

In geometric inequalities, the inequality that holds the most important place and which is most widely used is "Ravi Substitution"

Methodology: Suppose you are given a inequality dealing with the sides of a triangle, your first thought should be Rave Substitution. $$\cdots$$Suppose $a,b,c$ are sides of a triangle. According to ravi substitution, substitute $$a=x+y,b=y+z,c=z+x$$

$\mathcal{Example:}$Let $a,b,c$ be sides of a triangle. Prove that $$abc\geq(b+c-a)(a+b-c)(c+a-b)$$

$\mathcal{Solution:}$According to Ravi substitution, put $a=x+y,b=y+z,c=z+x$ , So, the inequality becomes; $$(x+y)(y+z)(z+x)\geq8xyz$$ Which follows trivially from AM-GM inequality.

$2] (IMO \: 1983;6)$ Let $a,b,c$ be sides of a triangle. Prove that $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0$$ This IMO problem, pretended to be hard, becomes too trivial by Ravi Substitution. Again, putting $a=x+y,b=y+z,c=z+x$ We obtain $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z$$

Which is very easy using Cauchy Schwartz, as $$(x+y+z)(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})\geq (x+y+z)^2$$

Trig substitutions in algebra

As the name suggests, we are going to talk about solving algebra problems by substituting trig functions or values. Why do we need this?

Sometimes the problem can be pretty ugly at first glance, and if you purely spamming pointless expansions, factorizations, squaring and so on, it will bring you to nowhere, instead, it gives an uglier look to the problem.

How do we prevent this? Let's look at an example.

Example 1. (This is used for training the 1996 USA IMO team)

Find a value of $x$ satisfying the equation $$5\sqrt{1-x}+\sqrt{1+x}=6x+8\sqrt{1-x^2}$$

What's your reaction when you first see this? Mentally booing the question? Feeling to give up? Everything's going to be okay with the trig substitutions.

Now, recall some basic formulas like $\sin^2 \theta + \cos^2\theta =1$, $\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$ and .

In fact, by substituting $x=\cos\theta$, one can immediately get $$\sqrt{1-x^2}=\sin\theta$$ and $$\sqrt{1\pm x}=\sqrt{1\pm\cos\theta}=\sqrt2\sqrt{\frac{1\pm\cos\theta}{2}}$$ Therefore the equation is transformed into $$5\sqrt2\Bigg(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\Bigg)=6\cos\theta+8\sin\theta$$

Time for a new tool! Given an expression of the form $a\cos\theta+b\sin\theta$, you will have $$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2}\Bigg(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta\Bigg)$$

Why? Consider this right triangle.

Therefore, recalling the summation formula, we have $$\begin{align*}a\cos\theta+b\sin\theta&=\sqrt{a^2+b^2}(\sin\alpha\cos\theta+\cos\alpha\sin\theta)\\&=\sqrt{a^2+b^2}\sin(\alpha+\theta)\end{align*}$$ In fact, we have $\sin x+\cos x=\sqrt2(x+\frac{\pi}{4})$. Applying this, the equation becomes (Note that $\sqrt{6^2+8^2}=10$) $$5\sqrt2\sqrt2\sin\Bigg(\frac{\theta}{2}+\frac{\pi}{4}\Bigg)=10\Bigg(\frac{3}{5}\cos\theta+\frac{4}{5}\sin\theta\Bigg)$$ Hence, when $\arctan\frac{3}{4}=\alpha$ (Do you know why?), we have $$\sin\Bigg(\frac{\theta}{2}+\frac{\pi}{4}\Bigg)=\sin(\alpha+\theta)$$ (Do you remember what we have done earlier to get this step?) Equating angles gives $\theta=\frac{\pi}{2}-2\alpha$. Thus, $$x=\cos\theta=\sin(2\alpha)=2(\sin\alpha)(\cos\alpha)=2\Bigg(\frac{3}{5}\Bigg)\Bigg(\frac{4}{5}\Bigg)=\frac{24}{25}$$. And We are done!

Before I end, I would like to give you a simple practice problem. Go and have a try. (The example above is just meant to tell you how one can solve equations using trigonometry, and don't expect you are going to use the steps above to solve the practice problem! Instead, consider some exact values of trig and C****e Rule!)

Practice 1.(Brilliant.org problem clone) Solve for $(x+y+z)^2$ if $$\begin{align*}x^2+y^2+xy&=64\\y^2+z^2+yz&=225\\x^2+z^2+xz&=289\end{align*}$$ and $$\begin{align*}x^2+y^2+xy&=100\\y^2+z^2+\sqrt{3}yz&=576\\x^2+z^2+\sqrt{3}xz&=676\end{align*}$$ (They are two separate sets of equations. Good luck and Happy Solving!)