0 reads0 ratings0 helpful
If (r=\sqrt{10}+\sqrt{2}), what is the value of (r^{2}-4\sqrt{5})?यदि (r=\sqrt{10}+\sqrt{2}), तो (r^{2}-4\sqrt{5}) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (12)(12)
Since (r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}), (r^{2}-4\sqrt{5}=12). In exams, subtract the radical middle term correctly.(r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}), इसलिए (r^{2}-4\sqrt{5}=12)। परीक्षा में करणी वाले मध्य पद को सही घटाएं।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
