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Algebra

no of solutions of (6-x)^4 +(8-x)^4=32

Profile image of shivali vij
14 Years agoGrade
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1 Answer

Profile image of SHAIK AASIF AHAMED
11 Years ago
Hello student,
Please find the answer to your question below
Given (6-x)^4 +(8-x)^4=32........(1)
Let y AM of 6-x and 8-x =7-x
so eqn(1) becomes (y+1)4+(y-1)4=32
on simplifying we get y4+6y2-15=0
let y2=u we get u2+6u-15=0
we get u=-3\pm2\sqrt{6}
so y=i\sqrt{2\sqrt{6}-3},-i\sqrt{2\sqrt{6}-3},-\sqrt{2\sqrt{6}-3},\sqrt{2\sqrt{6}-3}
so x=7+i\sqrt{2\sqrt{6}-3},7-i\sqrt{2\sqrt{6}-3},7+\sqrt{2\sqrt{6}-3},7-\sqrt{2\sqrt{6}-3}

Hence there are 4 solutions

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