Show that if a quadratic has rational coefficients and x+yz is a root, then x−yz must also be a root.

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Payton Cantrell · Accepted Answer

Maybe try using the following relationships of the roots and coefficients of a quadratic polynomial: Let a, b,&amp;nbsp;c∈Q.ax2+bx+c=a(x−x1)(x−x2)=a(x2−(x1+x2)+x1⋅x2)=ax2−a(x1+x2)x+a⋅x1⋅x2.From here we get the following system of equalities:{x1+x2=−ba&amp;nbsp;x1⋅x2=caNotice that if&amp;nbsp;x1=x+z, then&amp;nbsp;x1+x2=x+z+x2=−ba∈Q, then we can solve for&amp;nbsp;x2:x2=(−ba−x)−z=m−z,&amp;nbsp;&amp;nbsp;where&amp;nbsp;&amp;nbsp;m∈QNow let&#039;s use the second relationship we proved:x1⋅x2=(x+z)(m−z)=(xm−z)+(m−x)z=ca∈QIf&amp;nbsp;m−x≠0&amp;nbsp;we could solve for&amp;nbsp;z&amp;nbsp;to get that it must be rational, but this contradicts our assumption that it is irrational, hence&amp;nbsp;m=x, which implies&amp;nbsp;x2=m−z=x−z

Show that if a quadratic has rational coefficients and x

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