Question:
If x = √(5+2√6) / √(5-2√6), then determine the value of x² + 3x – 15?
यदि x = √(5+2√6) / √(5-2√6) है, तो x² + 3x – 15 का मान ज्ञात कीजिए?
Options:
- a) 50 + 12√6
- b) 40 + 23√6
- c) 49 + 26√6
- d) 35 + 15√6
- a) 50 + 12√6
- b) 40 + 23√6
- c) 49 + 26√6
- d) 35 + 15√6
correct answer : c)
Explanation:
Step 1: Simplify the expression for x.
We know that \( 5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3}\sqrt{2} = (\sqrt{3} + \sqrt{2})^2 \).
Similarly, \( 5 – 2\sqrt{6} = (\sqrt{3} – \sqrt{2})^2 \).
Thus, \( \sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2} \) and \( \sqrt{5 – 2\sqrt{6}} = \sqrt{3} – \sqrt{2} \).
Step 2: Express x and rationalize the denominator.
\( x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}} \)
Multiply numerator and denominator by \( (\sqrt{3} + \sqrt{2}) \):
\( x = \frac{(\sqrt{3} + \sqrt{2})^2}{3 – 2} = (\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6} \).
Step 3: Find x².
\( x^2 = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6} \).
Step 4: Substitute x and x² into the given expression \( x^2 + 3x – 15 \).
\( (49 + 20\sqrt{6}) + 3(5 + 2\sqrt{6}) – 15 \)
\( = 49 + 20\sqrt{6} + 15 + 6\sqrt{6} – 15 \)
\( = 49 + 26\sqrt{6} \).
चरण 1: x के व्यंजक को सरल करें。
हम जानते हैं कि \( 5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3}\sqrt{2} = (\sqrt{3} + \sqrt{2})^2 \)।
इसी प्रकार, \( 5 – 2\sqrt{6} = (\sqrt{3} – \sqrt{2})^2 \)।
इस प्रकार, \( \sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2} \) और \( \sqrt{5 – 2\sqrt{6}} = \sqrt{3} – \sqrt{2} \)।
चरण 2: x को व्यक्त करें और हर का परिमेयकरण करें。
\( x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}} \)
अंश और हर को \( (\sqrt{3} + \sqrt{2}) \) से गुणा करें:
\( x = \frac{(\sqrt{3} + \sqrt{2})^2}{3 – 2} = (\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6} \)।
चरण 3: x² ज्ञात करें。
\( x^2 = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6} \)।
चरण 4: दिए गए व्यंजक \( x^2 + 3x – 15 \) में x और x² का मान रखें。
\( (49 + 20\sqrt{6}) + 3(5 + 2\sqrt{6}) – 15 \)
\( = 49 + 20\sqrt{6} + 15 + 6\sqrt{6} – 15 \)
\( = 49 + 26\sqrt{6} \)।