Factorise x²-2√5x-3
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Answers
You do it the usual way.
The coefficient of the middle term is negative, but the constant term is positive. Hence, you must look for two numbers whose sum is 25–√25 and product is 33.
The product being rational, there must be some irrational conjugate multiplication happening here. If two irrational conjugates add to 25–√25, they must be 5–√±c5±c, where cc is a number we have to determine.
Their product is 33. Therefore, (5–√)2−c2=3(5)2−c2=3. Hence, the two numbers required are 5–√±2–√5±2.
How do I factorise x^2-2√5x+3?
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6 ANSWERS
Vishal Chandratreya
Vishal Chandratreya, former Assistant Engineer at Samsung Semiconductor India (2018)
Answered Jun 15, 2018
You do it the usual way.
The coefficient of the middle term is negative, but the constant term is positive. Hence, you must look for two numbers whose sum is 25–√25 and product is 33.
The product being rational, there must be some irrational conjugate multiplication happening here. If two irrational conjugates add to 25–√25, they must be 5–√±c5±c, where cc is a number we have to determine.
Their product is 33. Therefore, (5–√)2−c2=3(5)2−c2=3. Hence, the two numbers required are 5–√±2–√5±2.
Express the middle term and the constant in terms of these numbers.
x2−25–√x+3=x2−(5–√+2–√+5–√−2–√)x+(5–√+2–√)(5–√−2–√)x2−25x+3=x2−(5+2+5−2)x+(5+2)(5−2)
⟹x2−25–√x+3=x2−(5–√+2–√)x−(5–√−2–√)x+(5–√+2–√)(5–√−2–√)⟹x2−25x+3=x2−(5+2)x−(5−2)x+(5+2)(5−2)
⟹x2−25–√x+3=x[x−(5–√+2–√)]−(5–√−2–√)[x−(5–√+2–√)]⟹x2−25x+3=x[x−(5+2)]−(5−2)[x−(5+2)]
⟹x2−25–√x+3=[x−(5–√+2–√)][x−(5–√−2–√)]⟹x2−25x+3=[x−(5+2)][x−(5−2)]
Alternatively (and this is at worst), you could simply use the quadratic formula to determine its zeros, and then express the polynomial as a product of two linear factors.
−(−25–√)±(−25–√)2−4×1×3−−−−−−−−−−−−−−−−√2×1=5–√±2–√−(−25)±(−25)2−4×1×32×1=5±2
⟹x2−25–√x+3=[x−(5–√+2–√)][x−(5–√−2–√)]⟹x2−25x+3=[x−(5+2)][x−(5−2)]
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