Weekly Challenge 39: Symmetrically So

7077 /www/nrich/html/content/id/7077/ 1 2011-03-15T15:16:59 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/6267&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7074">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1274115048"> Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/7075">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7065&part=solution">Last week's solution</a> </li> </ul> <div> </div> <div>Make a substitution to find two exact real solutions to the equation $(x + 3)^4 + (x + 5) ^4 = 20.$</div> <div class="framework"> Did you know ... ? Frequently mathematicians spend their time stuck wondering how to solve equations or problems. One way of cracking a tough problem is to make a transformation to turn it into a more familiar form which allows the solution to proceed. Finding good substitutions or transformations is one of the more creative aspects of mathematics.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> In this case make the substitution $y=x+4$, which transforms the equation into $$ (y-1)^4+(y+1)^4=20. $$ This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'. If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur: $$ (y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20. $$ So, the odd factors cancel to give $$ 2y^4+12y^2-18=0 $$ This then becomes the simple equation $$ y^4+6y^2-9=0\,, $$ which is a quadratic equation in $y^2$ with solutions $$ y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2). $$ One of these solutions is positive; taking the square root gives two real values for $y$ as $$ y = \pm \sqrt{3\left(\sqrt{2}-1)\right)} $$ Therefore two real solutions to the equation are $$ x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)} $$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Try to solve this equation without multiplying it out. Notice a symmetry and make a substitution to a new variable and you can quickly solve this quartic by turning it into a quadratic... </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> In this case make the substitution $y=x+4$, which transforms the equation into $$ (y-1)^4+(y+1)^4=20. $$ This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'. If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur: $$ (y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20. $$ So, the odd factors cancel to give $$ 2y^4+12y^2-18=0 $$ This then becomes the simple equation $$ y^4+6y^2-9=0\,, $$ which is a quadratic equation in $y^2$ with solutions $$ y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2). $$ One of these solutions is positive; taking the square root gives two real values for $y$ as $$ y = \pm \sqrt{3\left(\sqrt{2}-1)\right)} $$ Therefore two real solutions to the equation are $$ x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)} $$ </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 39: Symmetrically So Exploit the symmetry and turn this quartic into a quadratic. Algebra Quadratic equations Advanced Algebra Polynomials