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Exercise 2.13 · Q19

Q.The number of roots of (x+3)4+(x+5)4=16(x+3)^4+(x+5)^4=16 is

(1) 44
(2) 22
(3) 33
(4) 00
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Step 1. Let u=x+4u=x+4, so x+3=u−1x+3=u-1 and x+5=u+1x+5=u+1. The equation becomes (u−1)4+(u+1)4=16(u-1)^4+(u+1)^4=16.

Step 2. Expand and add: (u−1)4+(u+1)4=2u4+12u2+2(u-1)^4+(u+1)^4=2u^4+12u^2+2, so 2u4+12u2+2=16⇒u4+6u2−7=02u^4+12u^2+2=16\Rightarrow u^4+6u^2-7=0.

Step 3. Let v=u2v=u^2: v2+6v−7=0⇒(v+7)(v−1)=0⇒v=1v^2+6v-7=0\Rightarrow(v+7)(v-1)=0\Rightarrow v=1 or v=−7v=-7. …

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