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Mathematics 16 Online
OpenStudy (anonymous):

what is the exact value of the equation sqrt of 52 - a sqrt 12 + sqrt of 117

OpenStudy (anonymous):

\[\sqrt{52} - \sqrt{13} + \sqrt{117}\]

OpenStudy (anonymous):

\[\sqrt[4]{13}\] \[\sqrt{13}\] \[\sqrt[2]{39}\] \[\sqrt[8]{39}\]

OpenStudy (vishweshshrimali5):

Notice that: 52 = 13 * 4 13 = 13 * 1 117 = 13 * 9

OpenStudy (anonymous):

yes

OpenStudy (vishweshshrimali5):

And if I am right, the options should be: (1) \(\large{4\sqrt{13}}\) (2) \(\large{\sqrt{13}}\) (3) \(\large{2\sqrt{13}}\) (4) \(\large{8\sqrt{13}}\)

OpenStudy (anonymous):

the options are above

OpenStudy (vishweshshrimali5):

Now, see: \[\large{\sqrt{117} = \sqrt{13 * 9} = \sqrt{13} * \sqrt{9} = \sqrt{13} * 3}\] Any doubt ?

OpenStudy (anonymous):

no doubt

OpenStudy (vishweshshrimali5):

Good... Similarly, can you write \(\large{\sqrt{52}}\) as \(\large{2*\sqrt{13}}\) ? Remember that 52 = 13 * 4

OpenStudy (anonymous):

ok

OpenStudy (vishweshshrimali5):

Great, so you have: \[\sqrt{52} - \sqrt{13} + \sqrt{117}\] \[= 2\sqrt{13} - \sqrt{13} + 3\sqrt{13}\] Now can you solve this?

OpenStudy (cwrw238):

@SuperMcMadrid5 - Note the the expression in your question is not an equation because it contains no equal sign

OpenStudy (anonymous):

it does not come with one

OpenStudy (vishweshshrimali5):

@SuperMcMadrid5 do you have some doubt ?

OpenStudy (anonymous):

no, i cant solve it though

OpenStudy (vishweshshrimali5):

Its okay lets see: \[= 2\sqrt{13} - \sqrt{13} + 3\sqrt{13}\] Did you get until this ?

OpenStudy (anonymous):

ok

OpenStudy (vishweshshrimali5):

Great! So, now: \[2\sqrt{13} - \sqrt{13} + 3\sqrt{13}\] \[=\sqrt{13}(2-1+3)\] Clear ?

OpenStudy (anonymous):

clear

OpenStudy (vishweshshrimali5):

Superb !! :) Now, what is the value of (2-1+3) ?

OpenStudy (anonymous):

4

OpenStudy (vishweshshrimali5):

Great going!! Now one final thing: \[=\sqrt{13}(2-1+3)\] \[= 4\sqrt{13}\] Now which option is this ?

OpenStudy (anonymous):

its A

OpenStudy (anonymous):

Thank you

OpenStudy (vishweshshrimali5):

Your welcome :)

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