How do you simplify #sqrt3 - 2sqrt3#? Algebra Radicals and Geometry Connections Addition and Subtraction of Radicals 1 Answer Don't Memorise Sep 4, 2015 #=color(blue)(-9# Explanation: #sqrt3 -2sqrt3# can be simplified by multiplying the expression by its the conjugate which #=color(blue)(sqrt3 +2sqrt3# =#(sqrt3 -2sqrt3 )* (color(blue)(sqrt3 +2sqrt3))# Now, we can apply the property : #(a-b)(a+b) = color(blue)(a^2-b^2# #(sqrt3 -2sqrt3 )* (color(blue)(sqrt3 +2sqrt3)) = (sqrt3)^2 - (2sqrt3)^2# #=3- 4*3 # #=3-12# #=color(blue)(-9# Answer link Related questions How do you add and subtract radicals? How is a radical considered a "like term"? How do you simplify #4\sqrt{3}+2\sqrt{12}#? How do you add #3""^3sqrt(2)+5""^3sqrt(16)#? How do you subtract #\sqrt{8x^3}-4x\sqrt{98x}#? How do you combine the radical #\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}#? How do you simplify #""^3sqrt{\frac{16x^5}{135y^4}}#? What is #sqrt(50)-sqrt(18)#? How do you add #3sqrt2+4sqrt2#? What is the square root of 50 + the square root of 8? See all questions in Addition and Subtraction of Radicals Impact of this question 2846 views around the world You can reuse this answer Creative Commons License