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Trig help? 1. Solve for x. sqrt 2x+3 = sqrt 6x-1 2. Solve for b. b^4/25=4
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1. \[\sqrt{2x+3} = \sqrt{6x-1}\] 2. \[\sqrt{b^4/25} = 4\]
for the first one square both sides... you'll get 2x+3=6x-1 ... can you finish the first one now ? :)
2x-6x=-1-3 -4x=-4 we multiply both sides by -1 4x=4 x=4/4 x=1
is it 1??
yes it is.. well done. For the second one I'll give you a hint... let's see what you can do :)
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\[\large \sqrt{\frac{b^4}{25}}=4\] \[\large \frac{\sqrt{b^4}}{\sqrt{25}}=4\] \[\large \frac{b^2}{5}=4\] can you do it now ? :)
is it +/- 2 √5 ??
yes it is... well done :)
Thank you so much!!! :)
My pleasure ..
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