Suppose x1,x2 are roots of the equation x=5−x2. Find the equation with roots 1(x1+1)3 and 1(x2+1)3.1) 125x2+16x=12) 125x2=16x+13) 125x2=12x+14) 125x2+12x=1

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Suppose x1,x2 are roots of the equation x=5−x2. Find the equation with roots 1(x1+1)3 and 1(x2+1)3.1) 125x2+16x=12) 125x2=16x+13) 125x2=12x+14) 125x2+12x=1

Charlie Haley · Accepted Answer

Using that:-&amp;nbsp;x2=−x+5-&amp;nbsp;x3=x⋅x2=−x2+5x=x−5+5x=6x−5Then:&amp;nbsp;y=1(x+1)3=1x3+3x2+3x+1=1(6x−5)+3(−x+5)+3x+1=16x+11&amp;nbsp;&amp;nbsp;⟺&amp;nbsp;6x=1y−11Substituting the latter in&amp;nbsp;36x2+36x−180=0&amp;nbsp;and multiplying by&amp;nbsp;y2&amp;nbsp;gives:(1−11y)2+6y(1−11y)−180y2=0;;;;⇔;;;;125y2+16y−1=0

Gia Edwards · Accepted Answer

As you can see the roots given for the new equation are of the format 1(x+1)3 hence you can use root transformation, either you directly transformx→1(x+1)3or you can do x→11+xand then calculate a bit, I&#039;m going to do using the second method as it reduces the problems and calculations involving irrationalities. Sayy=11+x⇒x=1−yyNow putting x on the original equation I get,1−yy=5−(1−yy)2⇒5y2+y−1=0Now, you can observe that the roots of this equation are 11+x1,11+x2. We see that the roots we want are just these roots raised to the power 3 Hence, sum of the roots we need is −(15)3+35×−15=−16125 similarly you can count the product of roots which will be −15 hence the equation we need is125x2+16x−1=0

Suppose \(\displaystyle{x}_{{{1}}},{x}_{{{2}}}\) are roots of the equation

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