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 $20$??

  • 1
    $\begingroup$ 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

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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.