Question

√(21-4√5+8√3-4√15) please solve this problem​

Answer 1

Step-by-step explanation:

Can you solve √ (21-4√5+8√3-4√15) here first square root is over all the equation?

Question: Can you solve √ (21-4√5+8√3-4√15) here first square root is over all the equation?

Solution:

21−45–√+83–√−415−−√−−−−−−−−−−−−−−−−−−−−√

The clue lies in the terms 5–√,3–√,15−−√Note: 15−−√=3–√∗5–√

Also, there is a constant term. One can conclude that expression when simpiefied must be of the form (a+b3–√+c5–√)

∴ 21−45–√+83–√−415−−√−−−−−−−−−−−−−−−−−−−−√=(a+b3–√+c5–√)(E01)

Squaring both sides:

⟹21−45–√+83–√−415−−√=(a+b3–√+c5–√)2

⟹21−45–√+83–√−415−−√=a2+3b2+5c2+2ab3–√+2bc15−−√+2ca5–√

Comparing the like terms on both sides:

15−−√ term: −4=2bc⟹bc=−2(E02)

5–√ term: −4=2ca⟹ca=−2(E03)

3–√ term: 8=2ab⟹ab=4(E04)

From (E02) and (E03):bcca=−2−2⟹a=b

From (E04):ab=4⟹a2=4⟹a=b=±2

From (E02):bc=−2⟹c=∓1

∴(E01):21−45–√+83–√−415−−√−−−−−−−−−−−−−−−−−−−−√=±(2+23–√−5–√)

Validation:

(2+23–√−5–√)2=22+(23–√)2+(5–√)2+(2⋅2⋅23–√)−(2⋅23–√⋅5–√)−(2⋅2⋅5–√)

=4+12+5+83–√−415−−√−45–√

=21+83–√−415−−√−45–√=Given expression. So, answer is good.