Weekly Challenge 10: Solve me!

7038 /www/nrich/html/content/id/7038/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/5876&part=">Try this next</a> </li> <li> <a href="http://en.wikipedia.org/wiki/Numerical_analysis">Read all about it</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1274115048">Ask NRICH</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1289816146">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/277&part=">Warm- up problem</a> </li> <li> <a href="http://nrich.maths.org/7040&part=solution">Last week's solution</a> </li> </ul> <div> Find a solution to this equation to 1 dp. $$2x^3+34 x^2+567x +8901=0$$ Are there any others? </div> <div class="framework"> Did you know ... ? Numerical solution of equations forms an important part of real-world mathematics and mathematics applied to science, where equations are often too complex to be solved exactly. Mathematicians have developed many advanced techniques for the numerical solution and exploration of equations.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Any sensible numerical method will lead to a solution $-16.3(2)$. In particular, an interval-halving method is efficient and simple to implement. You can make it a little quicker by choosing a sensible starting point: note that any solution would have to be negative, since all of the coefficients are positive; another moment of inspection will also show that the solution must lie between $-10$ and $-100$, giving you a sensible starting point for a computation. To determine whether there are any other solutions, note that the expression is a cubic and will therefore have either $1$ or $3$ real solutions. To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$, look at the turning points. Differentiating, we find that $$ \frac{dy}{dx} = 6x^2+68x+567 $$ The discriminiant of this quadratic is $68^2-4\times 6 \times 567 = -8984$. Since this is negative, there are no turning points and the cubic consequently only has $1$ real solution. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem draws together ideas of numerical solution of equations and calculus typically found at the start of a post-16 course. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Any sensible numerical method will lead to a solution $-16.3(2)$. In particular, an interval-halving method is efficient and simple to implement. You can make it a little quicker by choosing a sensible starting point: note that any solution would have to be negative, since all of the coefficients are positive; another moment of inspection will also show that the solution must lie between $-10$ and $-100$, giving you a sensible starting point for a computation. To determine whether there are any other solutions, note that the expression is a cubic and will therefore have either $1$ or $3$ real solutions. To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$, look at the turning points. Differentiating, we find that $$ \frac{dy}{dx} = 6x^2+68x+567 $$ The discriminiant of this quadratic is $68^2-4\times 6 \times 567 = -8984$. Since this is negative, there are no turning points and the cubic consequently only has $1$ real solution. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 10: Solve me! A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Numerical Methods A2 Secondary Mapping Document DisplayCabinet