find the value of a and b in the where 4+3√5÷4-3√5=a+b√5
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Anonymous:
3√5 ka whole square
ok thank you
wlcm
oo one more please... this 24
2×3×4√3
2ab
2×4×3√5
thank
Thanks :)
wlcm
Answered by
1
(4+3√5)/4-3√5
Rationalising the given expression by multiply 4+3√5 with the numerator as well as the denominator, we get
((4+3√5)^2)/((4)^2-(3√5)^2)
(16+45+24√5)/(16-45)
(61+24√5)/-29
-61/29+(-24√5/29)
Comoaring the abive exoression with a+b√5 we get
a=-61/29
b=-24/29
Rationalising the given expression by multiply 4+3√5 with the numerator as well as the denominator, we get
((4+3√5)^2)/((4)^2-(3√5)^2)
(16+45+24√5)/(16-45)
(61+24√5)/-29
-61/29+(-24√5/29)
Comoaring the abive exoression with a+b√5 we get
a=-61/29
b=-24/29
this 45 fom where we got
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