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Question

# Question 11 Find the values of a and b in each of the following (i)5+2√37+4√3=a−6√3 (ii)3−√53+2√5=a√5−1911 (iii)√2+√33√2−2√3=2−b√6 (iv)7+√57−√5−7−√57+√5=a+711√5b

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Solution

## (i) We have 5+2√37+4√3=a−6√3 For rationalizing the above equation, we multiply numerator and denominator of LHS by 7 - 4 √3, we get, 5+2√37+4√3×7−4√37−4√3=a−6√35(7−4√3)+2√3(7−4√3)72−(4√3)2=a−6√3 [Using identity,(a+b)(a−b)=a2−b2] ⇒35−20√3+14√3−2449−48=a−6√3⇒11−6√3=a−6√3⇒a=11 (ii) We have, 3−√53+2√5=a√5−1911 For rationalizing the above equation, we multiply numerator and denominator of LHS by 3−2√5, we get ⇒(3−√5)3+2√5×3−2√53−2√5=a√5−1911⇒3(3−2√5)−√5(3−2√5)(3)2−2√5)2=a√5−1911 [Using identity, (a−b)(a+b)=a2−b2] ⇒9−6√5−3√5+109−4×5=a√5−1911⇒19−9√59−20=a√5−1911⇒19−9√5−11=a√5−1911⇒9√511−1911=a√5−1911⇒9√511=a√5∴a=911 (iii) We have, √2+√33√2−2√3=2−b√6 For rationalising the above equation, we multiply numerator and denominator of LHS by 3√2+2√3, we get √2+√33√2−2√3×3√2+2√33√2+2√3=2−b√6⇒√2(3√2+2√3)+√3(3√2+2√3)(3√2)2−(2√3)2=2−b√6 [Using identity,(a−b)(a+b)=a2−b2] ⇒6+2√6+3√6+618−12=2−b√6⇒12+5√66=2−b√6⇒2+5√66=2−b√6⇒−b√6=5√66∴b=−56 (iv) We have, 7+√57−√5−7−√57+√5=a+711√5b ⇒(7+√5)2−(7−√5)2(7−√5)(7+√5)=a+711√5b⇒[72+(√5)2+2×7×√5]−(72+(√5)2−2×7×√5]72−(√5)2=a+711√5b ⎧⎪⎨⎪⎩Using identity,(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2and(a−b)(a+b)=a2−b2⎫⎪⎬⎪⎭ ⇒49+5+14√5−49−5+14√549−5=a+711√5b⇒28√544=a+711√5b⇒0+711√5=a+711√5b On comparing both sides, we get a = 0 and b = 1

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