Simplify : (√6 – 2√3)2
18 – 12√2
Simplify Radical Expression: Step-by-Step Solution
The question asks us to simplify the expression $(\sqrt{6} - 2\sqrt{3})^2$. This is a common type of problem involving squaring a binomial that contains square roots.
We can solve this by using the algebraic identity for squaring a binomial: $(a-b)^2 = a^2 - 2ab + b^2$.
In our expression, $(\sqrt{6} - 2\sqrt{3})^2$, we can identify:
- $a = \sqrt{6}$
- $b = 2\sqrt{3}$
Now, we substitute these values into the identity $(a-b)^2 = a^2 - 2ab + b^2$:
$(\sqrt{6} - 2\sqrt{3})^2 = (\sqrt{6})^2 - 2(\sqrt{6})(2\sqrt{3}) + (2\sqrt{3})^2$
Calculating Each Term
Let's calculate each part separately:
- Calculate $a^2 = (\sqrt{6})^2$:
Squaring a square root cancels out the root:
$(\sqrt{6})^2 = 6$ - Calculate $b^2 = (2\sqrt{3})^2$:
We square both the number outside the root and the root itself:
$(2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$ - Calculate $2ab = 2(\sqrt{6})(2\sqrt{3})$:
Multiply the numbers outside the roots and the numbers inside the roots:
$2(\sqrt{6})(2\sqrt{3}) = (2 \times 2) \times (\sqrt{6} \times \sqrt{3}) = 4 \times \sqrt{6 \times 3} = 4 \times \sqrt{18}$
Now, simplify $\sqrt{18}$. We look for perfect square factors of 18. $18 = 9 \times 2$, and 9 is a perfect square:
$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$
So, $2ab = 4 \times (3\sqrt{2}) = 12\sqrt{2}$
Combining the Simplified Terms
Now, substitute the simplified values of $a^2$, $b^2$, and $2ab$ back into the expression $a^2 - 2ab + b^2$:
$(\sqrt{6} - 2\sqrt{3})^2 = 6 - 12\sqrt{2} + 12$
Combine the constant terms (6 and 12):
$6 + 12 - 12\sqrt{2} = 18 - 12\sqrt{2}$
So, the simplified expression is $18 - 12\sqrt{2}$.
Let's verify this with the options provided.
| Option | Value |
|---|---|
| 1 | $18 + 2\sqrt{12}$ |
| 2 | $18 - 2\sqrt{12}$ |
| 3 | $18 + 12\sqrt{2}$ |
| 4 | $18 - 12\sqrt{2}$ |
Our calculated result $18 - 12\sqrt{2}$ matches Option 4.
Revision Table: Key Concepts in Simplifying Radical Expressions
| Concept | Description | Example |
|---|---|---|
| Squaring a square root | $(\sqrt{x})^2 = x$ for $x \ge 0$ | $(\sqrt{5})^2 = 5$ |
| Product of square roots | $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ for $a, b \ge 0$ | $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$ |
| Simplifying square roots | Factor the number inside the root to find perfect square factors. $\sqrt{a^2 b} = a\sqrt{b}$ | $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ |
| Binomial Square Identity | $(a-b)^2 = a^2 - 2ab + b^2$ | $(x-3)^2 = x^2 - 6x + 9$ |
Additional Information: Working with Radical Expressions
Radical expressions involve roots, most commonly square roots. Simplifying them often requires using properties of square roots and basic algebraic identities. When you have expressions like $(\sqrt{a} \pm \sqrt{b})^2$ or $(\sqrt{a} \pm c\sqrt{b})^2$, the binomial square identities are very useful.
- Remember to distribute the square: $(xy)^2 = x^2 y^2$. This applies to $(2\sqrt{3})^2 = 2^2 (\sqrt{3})^2$.
- Always simplify the radical part of your terms, like $\sqrt{18}$, to its simplest form, $3\sqrt{2}$, by factoring out any perfect squares.
- Combine like terms at the end. In this case, the constant terms (6 and 12) were combined, while the term with the square root ($12\sqrt{2}$) remained separate because it is not a like term.
Mastering these techniques is crucial for solving equations and simplifying expressions in algebra.
Important Questions from Identities
The coefficient of y in the expansion of (2y – 5) 3, is:
If x + y = 2 and \(\frac{1}{x}+\frac{1}{y}=\frac{18}{5}\) , then the value of (x 3+ y 3) is:
If x - y = 11 and \(\rm \frac{1}{x} - \frac{1}{y} = \frac{11}{24}\) then the value of x 3 - y 3 + x 2y 2 ?
If 2x 2- 8x - 1 = 0, then what is the value of \(\rm 8x^3 - \frac{1}{x^3}\) ?
If \(\rm x+ \frac{1}{x} = 4,\) then the value of \(\rm x^5 + \frac{1}{x^5}\) is:
