Finding the value of an expression

A favourite exam ques­tion gives you a surd value for $x$, such as $\displaystyle x = \frac{1}{2 - \sqrt{3}}$, and asks for the value of an expres­sion like $x^{3} - 2x^{2} - 7x + 5$. Every­thing you have learned so far comes together here: ratio­nal­is­ing, squar­ing with iden­ti­ties and col­lect­ing like surds. This les­son shows two meth­ods, and some neat short cuts for expres­sions such as $\displaystyle x + \frac{1}{x}$.

Method 1: work out each power#

Exam­ple 1#

If $\displaystyle x = \frac{1}{2 - \sqrt{3}}$, find the value of $x^{3} - 2x^{2} - 7x + 5$.

First ratio­nalise to get $x$ in a usable form:

$$\displaystyle x = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$$

Next find $x^{2}$ and $x^{3}$:

$$\begin{aligned} x^{2} &= (2 + \sqrt{3})^{2} = (2)^{2} + (\sqrt{3})^{2} + 2(2)(\sqrt{3}) \\ &= 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3} \\ x^{3} &= x^{2} \cdot x = (7 + 4\sqrt{3})(2 + \sqrt{3}) \\ &= 14 + 7\sqrt{3} + 8\sqrt{3} + 12 \\ &= 26 + 15\sqrt{3} \end{aligned}$$

Now sub­sti­tute:

$$\begin{aligned} x^{3} - 2x^{2} - 7x + 5 &= (26 + 15\sqrt{3}) - 2(7 + 4\sqrt{3}) - 7(2 + \sqrt{3}) + 5 \\ &= 26 + 15\sqrt{3} - 14 - 8\sqrt{3} - 14 - 7\sqrt{3} + 5 \\ &= 3 \end{aligned}$$

The surd terms can­cel: $15\sqrt{3} - 8\sqrt{3} - 7\sqrt{3} = 0$. The ratio­nal terms give $26 - 14 - 14 + 5 = 3$.

Method 2: find an equa­tion that x sat­is­fies#

Method 1 needs a lot of mul­ti­ply­ing. There is a clev­erer way. Since $x = 2 + \sqrt{3}$,

$$\begin{aligned} x - 2 &= \sqrt{3} \\ (x - 2)^{2} &= (\sqrt{3})^{2} \\ x^{2} - 4x + 4 &= 3 \\ x^{2} - 4x + 1 &= 0 \end{aligned}$$

Squar­ing removed the surd, so $x$ sat­is­fies the sim­ple equa­tion $x^{2} - 4x + 1 = 0$.

Now divide $x^{3} - 2x^{2} - 7x + 5$ by $x^{2} - 4x + 1$, using long divi­sion as in the fac­tori­sa­tion course:

• $x^{3} \div x^{2} = x$; sub­tract $x(x^{2} - 4x + 1) = x^{3} - 4x^{2} + x$; this leaves $2x^{2} - 8x + 5$.
• $2x^{2} \div x^{2} = 2$; sub­tract $2(x^{2} - 4x + 1) = 2x^{2} - 8x + 2$; this leaves $3$.

So the quo­tient is $x + 2$ and the remain­der is $3$:

$$\begin{aligned} x^{3} - 2x^{2} - 7x + 5 &= (x^{2} - 4x + 1)(x + 2) + 3 \\ &= 0 \times (x + 2) + 3 \\ &= 3 \end{aligned}$$

The same answer, with no surd arith­metic at all after the first step.

Short cuts for x and 1 over x#

Many ques­tions ask for $\displaystyle x + \frac{1}{x}$ or $\displaystyle x^{2} + \frac{1}{x^{2}}$. Two facts help.

• If $x = a + \sqrt{b}$ and $a^{2} - b = 1$, then $\displaystyle \frac{1}{x}$ is sim­ply the con­ju­gate $a - \sqrt{b}$, because $(a + \sqrt{b})(a - \sqrt{b}) = 1$.
• From the iden­tity $(x + y)^{2} = x^{2} + y^{2} + 2xy$ with $\displaystyle y = \frac{1}{x}$:

$$\displaystyle x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} - 2$$

Exam­ple 2#

If $x = 3 + 2\sqrt{2}$, find $\displaystyle x + \frac{1}{x}$ and $\displaystyle x^{2} + \frac{1}{x^{2}}$.

Since $(3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1$,

$$\displaystyle \frac{1}{x} = 3 - 2\sqrt{2}$$

So

$$\displaystyle x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$$

and

$$\displaystyle x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} - 2 = 36 - 2 = 34$$

Exam­ple 3: a cube root#

If $x = \sqrt[3]{2 + \sqrt{3}}$, find $\displaystyle x^{3} + \frac{1}{x^{3}}$.

Cub­ing both sides, $x^{3} = 2 + \sqrt{3}$. Since $(2 + \sqrt{3})(2 - \sqrt{3}) = 1$, we have $\displaystyle \frac{1}{x^{3}} = 2 - \sqrt{3}$. So

$$\displaystyle x^{3} + \frac{1}{x^{3}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$$

The cube root looked fright­en­ing, but the ques­tion only needed $x^{3}$.

Exam­ple 4: two related val­ues#

If $\displaystyle a = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ and $\displaystyle b = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$, find $a^{2} + ab + b^{2}$.

Ratio­nalise each:

$$\displaystyle a = \frac{(\sqrt{3} + \sqrt{2})^{2}}{3 - 2} = 5 + 2\sqrt{6}, \qquad b = \frac{(\sqrt{3} - \sqrt{2})^{2}}{3 - 2} = 5 - 2\sqrt{6}$$

Then $a + b = 10$ and $ab = 25 - 24 = 1$. Using $a^{2} + b^{2} = (a + b)^{2} - 2ab$:

$$a^{2} + ab + b^{2} = (a + b)^{2} - 2ab + ab = (a + b)^{2} - ab = 100 - 1 = 99$$

Work­ing with $a + b$ and $ab$ avoids squar­ing the surds at all.

Prac­tice#

1. If $x = 2 + \sqrt{3}$, find $\displaystyle x + \frac{1}{x}$.
2. If $x = 2 + \sqrt{3}$, find $\displaystyle x^{2} + \frac{1}{x^{2}}$.
3. If $x = \sqrt{5} + 2$, find $\displaystyle x - \frac{1}{x}$.
4. If $x = 4 - \sqrt{15}$, find $\displaystyle x^{2} + \frac{1}{x^{2}}$.
5. If $\displaystyle x = \frac{1}{3 - 2\sqrt{2}}$, find the value of $x^{2} - 6x + 1$.
6. If $x = 1 + \sqrt{2}$, find the value of $x^{3} - 2x^{2} - x + 3$.
7. If $\displaystyle a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$ and $\displaystyle b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}$, find $a^{2} + b^{2}$.
8. If $x = \sqrt[3]{3 + 2\sqrt{2}}$, find $\displaystyle x^{3} + \frac{1}{x^{3}}$.

Answers#