find the value of a in the following =6/3 root 2 - 2 root 3 = 3 root 2 - a root 3
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Answered by
346
6/(3√2-2√3)
Here we have to rationalise the denominator,
Rationalising factor=(3√2+2√3)
=6/(3√2-2√3)×(3√2+2√3)/(3√2+2√3)
=6(3√2+2√3)/(3√2-2√3)(3√2+2√3)
=(18√2+12√3)/[(3√2)²-(2√3)²]
=(18√2+12√3)/(9(2)-4(3))
=(18√2+12√3)/18-12
=(18√2+12√3)/6
=(18√2/6)+(12√3/6)
=3√2+2√3
=3√2-(-2√3)
Compare this with 3√2-a√3
3√2-(-2√3)=3√2-a√3
Therefore, a= -2
Hope it helps...
Here we have to rationalise the denominator,
Rationalising factor=(3√2+2√3)
=6/(3√2-2√3)×(3√2+2√3)/(3√2+2√3)
=6(3√2+2√3)/(3√2-2√3)(3√2+2√3)
=(18√2+12√3)/[(3√2)²-(2√3)²]
=(18√2+12√3)/(9(2)-4(3))
=(18√2+12√3)/18-12
=(18√2+12√3)/6
=(18√2/6)+(12√3/6)
=3√2+2√3
=3√2-(-2√3)
Compare this with 3√2-a√3
3√2-(-2√3)=3√2-a√3
Therefore, a= -2
Hope it helps...
sivaprasath:
the denominator is already rationalised,3√2+2√3 is completed,,..??
Answered by
183
Answer is shown in picture:
a=-2
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