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Question Number 104588 by  M±th+et+s last updated on 22/Jul/20

solve in R ((96x−24)/(12x+5))=(√(−144x^2 +72x+7))

solveinR96x−2412x+5=−144x2+72x+7

Commented by  M±th+et+s last updated on 22/Jul/20

thanks both for solutions

thanksbothforsolutions

Answered by ajfour last updated on 22/Jul/20

8−(((64)/(12x+5)))=(√(16−(12x−3)^2 )) let 12x−3=t with −4≤t≤4 ⇒ 8−((64)/(t+8))=(√(16−t^2 )) ⇒ 4−(√(16−t^2 ))=((64)/(t+8))−4 ....(i) ⇒ (t^2 /(4+(√(16−t^2 )))) = ((64)/(t+8))−4 ⇒ 4+(√(16−t^2 )) = (t^2 /4)(((t+8)/(8−t))) ...(ii) Adding (i) & (ii) 8=(t^2 /4)(((t+8)/(8−t)))+4(((8−t)/(t+8))) Now let ((8−t)/(t+8)) = s ⇒ t=8(((1−s)/(s+1))) ⇒ 8=16(((1−s)/(s+1)))^2 ((1/s))+4s ⇒ 2s(s+1)^2 =4(1−s)^2 +s^2 (s+1)^2 ⇒ s(s+1)^2 (2−s)=4(1−s)^2 ⇒ (s+1)^2 [1−(s−1)^2 ]=4(1−s)^2 ⇒ (1/((1−s)^2 ))−(4/((1+s)^2 )) = 1 ......

8−(6412x+5)=16−(12x−3)2let12x−3=twith−4⩽t⩽4⇒8−64t+8=16−t2⇒4−16−t2=64t+8−4....(i)⇒t24+16−t2=64t+8−4⇒4+16−t2=t24(t+88−t)...(ii)Adding(i)&(ii)8=t24(t+88−t)+4(8−tt+8)Nowlet8−tt+8=s⇒t=8(1−ss+1)⇒8=16(1−ss+1)2(1s)+4s⇒2s(s+1)2=4(1−s)2+s2(s+1)2⇒s(s+1)2(2−s)=4(1−s)2⇒(s+1)2[1−(s−1)2]=4(1−s)2⇒1(1−s)2−4(1+s)2=1......

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