http://sgforums.com/forums/2297/topics/322601 en-US 60 <p>Another easy method to solve for k+1 for the complex roots:</p> <p>You draw the Im-Real axis. You can also see that one of the roots is cube root of 1.5, which is a real number.</p> <p>Then you rotate the point around the origin by 120 degrees (360/3)</p> <p>one of the complex roots of k+1 would be:<br /> - 1.145 sin 30 + 1.145 cos 30 = -0.5725 + 0.9916i</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>I think both of the above methods would be easier than using and understanding theory of equations</p> Thu, 03 Jul 2008 15:25:26 +0800 sgforums.com:2297:322601:8214445 eagle http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by jiaxing2:</div> <div class="quote_body"> <p>there should be 2 more roots which are complex numbers...</p> <p>can be solved by theory of eqn</p> <p>sum of the 3 roots = -3</p> <p>product of 3 roots = 0.5</p> <p>since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )</p> <p>let k be the real root.&nbsp; 2a + k = -3</p> <p>solving a = -&nbsp;[ 1 + 0.5(1.5)^1/3 ]</p> <p>similarly ( a^2 + b^2&nbsp;) * k =0.5 (product of 3 roots)</p> <p>go solve for b...and u get ur 2 complex roots</p> <p>for ur info only...unless this qn carries&nbsp;alot of&nbsp;marks...i dun think u are really required to find&nbsp;out what the complex roots are...&nbsp;</p> </div> </blockquote> <p>No need do so hard to solve the complex roots also</p> <p>Simplified version</p> <p>(k+1)^3 - (cube root[1.5])^3 = 0</p> <p>you can use the formula x^3 - y^3 = (x^2 + xy + y^2)(x-y)<br /> where x = k+1 and y = cube root of 1.5</p> <p>Then you can do the normal use equation to solve for quadratic equations method to determine the final complex roots, in which your discriminant should be -ve</p> Thu, 03 Jul 2008 15:20:46 +0800 sgforums.com:2297:322601:8214436 eagle http://sgforums.com/forums/2297/topics/322601 <p>there should be 2 more roots which are complex numbers...</p> <p>can be solved by theory of eqn</p> <p>sum of the 3 roots = -3</p> <p>product of 3 roots = 0.5</p> <p>since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )</p> <p>let k be the real root.&nbsp; 2a + k = -3</p> <p>solving a = -&nbsp;[ 1 + 0.5(1.5)^1/3 ]</p> <p>similarly ( a^2 + b^2&nbsp;) * k =0.5 (product of 3 roots)</p> <p>go solve for b...and u get ur 2 complex roots</p> <p>for ur info only...unless this qn carries&nbsp;alot of&nbsp;marks...i dun think u are really required to find&nbsp;out what the complex roots are...&nbsp;</p> Wed, 02 Jul 2008 21:10:51 +0800 sgforums.com:2297:322601:8212865 jiaxing2 http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by ^tamago^:</div> <div class="quote_body"> <p><br /> yeah, but he want solve cubic. :(</p> </div> </blockquote> <p><br /> haha doesnt matter as long as i get the answer right</p> Mon, 30 Jun 2008 00:30:34 +0800 sgforums.com:2297:322601:8206285 brokenluv http://sgforums.com/forums/2297/topics/322601 <p>This one not function meh?</p> Mon, 30 Jun 2008 00:22:46 +0800 sgforums.com:2297:322601:8206252 jay_rocks http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by eagle:</div> <div class="quote_body"> <p>got simpler method</p> <p>1.5 = (1+k)³<br /> cube root(1.5) = (k+1)</p> <p>k = cube root(1.5) -1 = 0.145</p> </div> </blockquote> <p><br /> yeah, but he want solve cubic. :(</p> Mon, 30 Jun 2008 00:16:35 +0800 sgforums.com:2297:322601:8206220 ^tamago^ http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by eagle:</div> <div class="quote_body"> <p>got simpler method</p> <p>1.5 = (1+k)³<br /> cube root(1.5) = (k+1)</p> <p>k = cube root(1.5) -1 = 0.145</p> </div> </blockquote> <p><br /> wow thx! i totally forgot my amath after i left sec school</p> Mon, 30 Jun 2008 00:03:05 +0800 sgforums.com:2297:322601:8206158 brokenluv http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by ^tamago^:</div> <div class="quote_body"> <p>1.5 = (1+k)³<br /> 3 = 2(k³+3k²+3k+1)<br /> 2k³+6k²+6k-1=0<br /> k = 0.1447 ≈ 0.145 (3.s.f)</p> </div> </blockquote> <p>got simpler method</p> <p>1.5 = (1+k)³<br /> cube root(1.5) = (k+1)</p> <p>k = cube root(1.5) -1 = 0.145</p> Sun, 29 Jun 2008 23:59:54 +0800 sgforums.com:2297:322601:8206138 eagle http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by ^tamago^:</div> <div class="quote_body"> <p>2k³+6k²+6k-1=0<br /> <br /> is no different from<br /> <br /> k³+3k²+3k-½=0<br /> <br /> which you have gotten. now, this equation cannot be solved directly.</p> </div> </blockquote> <p><br /> erm its 2k^2, not 3k^2</p> Sun, 29 Jun 2008 23:58:35 +0800 sgforums.com:2297:322601:8206130 brokenluv http://sgforums.com/forums/2297/topics/322601 <p>2k³+6k²+6k-1=0<br /> <br /> is no different from<br /> <br /> k³+3k²+3k-½=0<br /> <br /> which you have gotten. now, this equation cannot be solved directly.</p> Sun, 29 Jun 2008 23:55:37 +0800 sgforums.com:2297:322601:8206121 ^tamago^ http://sgforums.com/forums/2297/topics/322601 <blockquote> <div class="quote_from">Originally posted by ^tamago^:</div> <div class="quote_body"> <p>1.5 = (1+k)³<br /> 3 = 2(k³+3k²+3k+1)<br /> 2k³+6k²+6k-1=0<br /> k = 0.1447 ≈ 0.145 (3.s.f)</p> </div> </blockquote> <p><br /> i dont understand this step</p> Sun, 29 Jun 2008 23:52:52 +0800 sgforums.com:2297:322601:8206101 brokenluv http://sgforums.com/forums/2297/topics/322601 <p><img src= "http://img241.imageshack.us/img241/503/picture1py7.png" alt= "image" /></p> Sun, 29 Jun 2008 23:49:22 +0800 sgforums.com:2297:322601:8206077 ^tamago^ http://sgforums.com/forums/2297/topics/322601 <p>1.5 = (1+k)³<br /> 3 = 2(k³+3k²+3k+1)<br /> 2k³+6k²+6k-1=0<br /> k = 0.1447 ≈ 0.145 (3.s.f)</p> Sun, 29 Jun 2008 23:45:42 +0800 sgforums.com:2297:322601:8206065 ^tamago^ http://sgforums.com/forums/2297/topics/322601 <p>1.5 = ( 1 + k ) ^3<br /> find k.<br /> <br /> i got to this step : k^3 + 2k^2 + 3k - 0.5 = 0<br /> and then i got stuck. please help thx!</p> Sun, 29 Jun 2008 23:34:58 +0800 sgforums.com:2297:322601:8206013 brokenluv http://sgforums.com/forums/2297/topics/322601