Like this. Factor inside the radical.
3sqrt(2^2*2^2 *5)
as you can see....4 *4 = 16 * 5 = 80, so you can bring both 2,s out and multiply them against the 3
2*2*3sqrt(5)
= 12sqrt(5)
simplified
2*sqrt(75) = 2*sqrt(25*3) = 2*sqrt(25)*sqrt(3) = 2*5*sqrt(3) = 10*sqrt(3)
3 sqrt (3)Put the radical where the "sqrt" is with the second 3 inside.
In order to work with radicals (adding or subtracting) they have to have the same radical expression. ex: 2 rad 3 + 4 rad 3 = 6 rad 3. If it helps, change the similar radicals to a variable. 2 rad 3 + 4 rad 3 2 x + 4 x = 6x and then substitute your radical back in for the variable. 6 rad 3 ---- rad 12 - 3 rad 3 Let's simplify radical 12. First, factor 12 into a perfect square, and a not perfect square.) Hint: we want it to be radical 3 to work with it... 3 times what is 12? rad 12 = sqrt(12) = sqrt(4*3) sqrt(4*3) = sqrt(4)*sqrt(3) sqrt(4) is just 2! We bring this to the "outside" of the radical. sqrt(4)*sqrt(3) = 2 sqrt(3) or 2 rad 3 rad 12 - 3 rad 3 2 rad 3 - 3 rad 3 (or 2x - 3x if it helps) We're left with: -1 rad 3, or just - rad 3 (negative radical 3)
sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3
sqrt(363) = sqrt(121*3) = sqrt(121)*sqrt(3) = 11*sqrt(3)
sqrt(27) = 3*sqrt(3).
2*sqrt(75) = 2*sqrt(25*3) = 2*sqrt(25)*sqrt(3) = 2*5*sqrt(3) = 10*sqrt(3)
Factorise 12.Replace each pair appearing in this factorisation as by the same number outside the radical and then put everything under the radical sign. sqrt(12) = sqrt(2*2*3) = 2*sqrt(3)
To simplify the expression (3\sqrt{75} + \sqrt{27}), we start by breaking down the radicals. First, ( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} ), so ( 3\sqrt{75} = 3 \times 5\sqrt{3} = 15\sqrt{3} ). Next, ( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} ). Now we can combine the two terms: ( 15\sqrt{3} + 3\sqrt{3} = (15 + 3)\sqrt{3} = 18\sqrt{3} ). Thus, ( 3\sqrt{75} + \sqrt{27} = 18\sqrt{3} ).
You can rationalize the denominator by multiplying this fraction by a fractional form of one in radical form. 3/sqrt(2) * sqrt(2)/sqrt(2) = 3sqrt(2)/2 ----------------
To simplify ( 3 \sqrt{432} ), first factor ( 432 ) into its prime factors: ( 432 = 2^4 \times 3^3 ). Then, rewrite the square root: ( \sqrt{432} = \sqrt{2^4 \times 3^3} = \sqrt{(2^2)^2 \times (3^1)^2 \times 3} = 2^2 \times 3 \times \sqrt{3} = 12\sqrt{3} ). Finally, multiply by ( 3 ): ( 3 \sqrt{432} = 3 \times 12\sqrt{3} = 36\sqrt{3} ).
sqrt(3) x sqrt(15) = sqrt( 3 x 15 ) = sqrt( 45 ) = sqrt( 9 x 5 ) = 3 sqrt(5)
To simplify ( \sqrt{18} + \sqrt{32} ), we first break down each square root. ( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} ) and ( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} ). Combining these gives ( 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2} ). Thus, ( \sqrt{18} + \sqrt{32} = 7\sqrt{2} ).
If we factor 60, we get 60 = 2 * 2 * 3 * 5 So sqrt ( 60 ) = sqrt (2 * 2 * 3 * 5) = sqrt (2 * 2) * sqrt (3 * 5) = 4 * sqrt( 15 ) ____________________________________________________________ actually if you do by just using the simpler way you should get 2 radical 15 since , 60 has the closest perfect square of 4 and non perfect of 15 this is how you should solve radical 60 = radical 4 times radical 15 radical 60 = simplify the 4 to 2 and leave radical 15 So , radical 60 = 2 radical 15 not what is above .
The SQRT(18) in radical form is 3*SQRT(2), where SQRT represents the radical symbol.
Yes. So 3*sqrt(2) + 5*sqrt(2) = 8*sqrt(2) But you may need to simplify terms eg 3*sqrt(2) + 5*sqrt(8) = 3*sqrt(2) + 10*sqrt(2) = 13*sqrt(2)
3 sqrt(6) x sqrt(6) = 18