JEE Mathematics: Indices & Rationalization (Lecture 03)
Topic 1: Rationalization of Surds & Simplification
- Rationalization is the process of eliminating surds (square roots, cube roots, etc.) from the denominator of a fraction.
- To rationalize a binomial denominator like $a + b\sqrt{x}$, multiply both the numerator and the denominator by its conjugate, $a – b\sqrt{x}$.
- Use the algebraic identity $(x + y)(x – y) = x^2 – y^2$ to simplify the denominator.
- Always simplify the square roots of large numbers by factoring out perfect squares before performing operations (e.g., $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$).
- સંમેયીકરણ એ અપૂર્ણાંકના છેદમાંથી અસંમેય સંખ્યાઓ (વર્ગમૂળ, ઘનમૂળ વગેરે) દૂર કરવાની પ્રક્રિયા છે.
- $a + b\sqrt{x}$ જેવા દ્વિપદી છેદનું સંમેયીકરણ કરવા માટે, અંશ અને છેદ બંનેને તેની અનુબદ્ધ કરણી $a – b\sqrt{x}$ વડે ગુણો.
- છેદનું સાદુંરૂપ આપવા માટે બૈજિક નિત્યસમ $(x + y)(x – y) = x^2 – y^2$ નો ઉપયોગ કરો.
- ગણતરી કરતા પહેલા હંમેશા મોટી સંખ્યાઓના વર્ગમૂળમાંથી પૂર્ણ વર્ગ સામાન્ય કાઢીને સાદુંરૂપ આપો (દા.ત., $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$).
Rationalize the denominator of $\frac{3}{2 + \sqrt{3}}$.
$\frac{3}{2 + \sqrt{3}}$ ના છેદનું સંમેયીકરણ કરો.
Solution:
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $2 – \sqrt{3}$.
$$ \frac{3}{2 + \sqrt{3}} \times \frac{2 – \sqrt{3}}{2 – \sqrt{3}} $$
Apply the difference of squares formula $(a+b)(a-b) = a^2 – b^2$ in the denominator:
$$ \frac{3(2 – \sqrt{3})}{(2)^2 – (\sqrt{3})^2} $$
$$ \frac{6 – 3\sqrt{3}}{4 – 3} $$
$$ \frac{6 – 3\sqrt{3}}{1} = 6 – 3\sqrt{3} $$
If $\frac{4+3\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}$ then find rational numbers a and b.
જો $\frac{4+3\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}$ તો સંમેય સંખ્યા a અને b મેળવો.
Solution:
We need to rationalize the denominator by multiplying the numerator and denominator by the conjugate of $4 – 3\sqrt{5}$, which is $4 + 3\sqrt{5}$.
$$ \frac{4+3\sqrt{5}}{4-3\sqrt{5}} \times \frac{4+3\sqrt{5}}{4+3\sqrt{5}} $$
Expand the numerator using $(x+y)^2 = x^2 + 2xy + y^2$:
$$ (4+3\sqrt{5})^2 = (4)^2 + 2(4)(3\sqrt{5}) + (3\sqrt{5})^2 $$
$$ 16 + 24\sqrt{5} + 45 = 61 + 24\sqrt{5} $$
Expand the denominator using $(x-y)(x+y) = x^2 – y^2$:
$$ (4)^2 – (3\sqrt{5})^2 = 16 – 45 = -29 $$
Now, put the numerator and denominator back together:
$$ \frac{61 + 24\sqrt{5}}{-29} = -\frac{61}{29} – \frac{24}{29}\sqrt{5} $$
Comparing this with $a + b\sqrt{5}$, we get:
$$ a = -\frac{61}{29}, \quad b = -\frac{24}{29} $$
Simplify: $\frac{(5\sqrt{3}+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt{2}}$
સાદુંરૂપ આપો: $\frac{(5\sqrt{3}+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt{2}}$
Solution:
First, simplify the individual surds by finding perfect square factors:
$$ \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} $$
$$ \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} $$
$$ \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} $$
Substitute these back into the given expression:
$$ \frac{(5\sqrt{3} + 5\sqrt{2})(5 – 2\sqrt{6})}{5\sqrt{3} – 5\sqrt{2}} $$
Factor out the common term $5$ from the first bracket in the numerator and the entire denominator:
$$ \frac{5(\sqrt{3} + \sqrt{2})(5 – 2\sqrt{6})}{5(\sqrt{3} – \sqrt{2})} $$
Cancel the $5$ from the numerator and denominator:
$$ \frac{(\sqrt{3} + \sqrt{2})(5 – 2\sqrt{6})}{\sqrt{3} – \sqrt{2}} $$
Now, rationalize the fraction part $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}}$ by multiplying the numerator and denominator by its conjugate $(\sqrt{3} + \sqrt{2})$:
$$ \frac{(\sqrt{3} + \sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} – \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 – (\sqrt{2})^2} $$
$$ \frac{3 + 2\sqrt{6} + 2}{3 – 2} = \frac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6} $$
Finally, multiply this result with the remaining term $(5 – 2\sqrt{6})$:
$$ (5 + 2\sqrt{6})(5 – 2\sqrt{6}) $$
Apply the difference of squares identity $(a+b)(a-b) = a^2 – b^2$:
$$ (5)^2 – (2\sqrt{6})^2 = 25 – 4(6) = 25 – 24 = 1 $$
The final simplified value is $1$.