Solvinga(a+1)+b(b+1)=12, (a+1)(b+1)=4is there any trick?

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Solvinga(a+1)+b(b+1)=12, (a+1)(b+1)=4is there any trick?

2sze1c1se3nh · Accepted Answer

Step 1Let u=a+1, u=b+1. The equations become(u−1)u+(v−1)v=12, uv=4Thusuv+(u−1)u+uv+(v−1)v=u(v+u−1)+v(u+v−1)=(u+v)(u+v−1)=20Let x=u+v and we have a quadratic, which yields u+v=5 or -4Solve the rest using this info and uv=4, which is just another quadratic.Solving the quartic doesn&#039;t seem too bad either, by substitution:(u−1)u+4u(4u−1)=12u2+16u2−u−4u=12Note that (u+4u)2=u2+8+16u2 Hence we can rewrite the above into:(u+4u)2−(u+4u)=20which, of course, is a quadratic equation with u+4u=5 or -4, and thus we arrive at essentially the same conclusion as the above (slightly unnatural) method.

amisayq6t · Accepted Answer

Step 1Expressed in elementary symmetric polynomialss=a+b, p=abthe equations become1) s2−2p+s=122) p+s=3To get rid of p, combine (1)+2(2):s2+3s=18So either s=3 (and p=0) or s=−6 (and p=9). At any rate, a,b are the roots of x2−sx+p, so the solutions are(a,b)=(0,3) or (3,0) or (−3,−3).

Solving a(a + 1) + b(b + 1) = 12,\ (a+

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