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Okay the question is to solve using quadratic formula and the question is $10r^2-6=0$ and after using the $-b \pm \frac{\sqrt{ b^2-4ac}}{2a}$ I got $-0 \pm 0^2-\frac{\sqrt{240}}{20}$...and I don't know where to go after that...i thought about taking square root but than i get $15$ something $20$??

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    $\begingroup$ Well, is there not a simpler way to solve the question? $10r^2=6$... $\endgroup$ – Feyre Sep 8 '15 at 21:27
  • $\begingroup$ As @Feyre said, it is simple solution. $r=\pm \sqrt{\frac 35}$ $\endgroup$ – user249332 Sep 8 '15 at 21:29
  • $\begingroup$ Just write out the prime factorization of $240$. You will see that it can be factored as $16\times 15$ $\endgroup$ – John Joy Sep 8 '15 at 22:48
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If you're required to use quadratic formula, you just need to simplify:

$\pm \frac{\sqrt{240}}{20} = \pm\frac{\sqrt{240}}{\sqrt{400}} =\pm \sqrt{\frac{240}{400}} = \pm\sqrt{\frac{3}{5}}$

Otherwise, as Feyre pointed out, there's no reason to use the quadratic formula and you can just solve it directly

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  • $\begingroup$ yea i had to use quadratic formula....thanks a lot for the answer but the answer in the answer sheet is ± square root of 15 divided by 5, but the 5 is not inside the square root...15 is the only one inside it....and also it im sorry but i had a error in the question... $\endgroup$ – MATH ASKER Sep 8 '15 at 21:54
  • $\begingroup$ @MATHASKER: $\frac{\sqrt{15}}5 = \frac{\sqrt{15}}{\sqrt{25}} = \sqrt{\frac{15}{25}} = \sqrt{\frac 35}$ $\endgroup$ – Peter Phipps Sep 8 '15 at 23:11
  • $\begingroup$ Most math classes will require reductions of all radicals, fractions, and rationalized denominators. So the reduction you'd see would be: $\frac {\sqrt{240}}{20} = \frac {\sqrt{16*15}}{20} = \frac {4 \sqrt{15}} {4*5} = \frac {\sqrt{15}}{5}$. $\endgroup$ – Daniel R. Collins Sep 9 '15 at 3:59

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