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How do you solve this equation using the square root property 13 Aug 2015 · How do you solve this equation using the square root property #(x + 3)^2 = 49#? Algebra Quadratic Equations and Functions Use Square Roots to Solve Quadratic Equations 1 Answer
How do you simplify #sqrt(49)#? - Socratic 14 May 2018 · sqrt49=7 As 7*7=49 it follows that 7 is the square root of 49. Therefore sqrt49=7 Since we don't have +- in front of the square root sign, we are only asked for the positive value.
How do you simplify sqrt98? - Socratic 3 Feb 2016 · sqrt(98) = 7sqrt(2) We will use the following properties: For a, b >= 0 sqrt(ab) = sqrt(a)sqrt(b) -sqrt(a^2) = a Then, applying this, sqrt(98) = sqrt(49*2) =sqrt(49 ...
How do you simplify 2 square root -49? - Socratic 12 Jan 2017 · If we use real numbers, this problem has no solution since there is no real number that is squared #- 49# and therefore, there is no square root of #- 49# in #RR#. But if we expand the set of numbers we use and use complex numbers #CC#, we have that the imaginary number #i# is: #i = sqrt {- 1}#.
How do you simplify #sqrt(49-x^2)#? - Socratic 31 Jan 2016 · This expression cannot be simplified, but it can be re-expressed: sqrt(49-x^2) = sqrt(7-x)sqrt(7+x) If a >= 0 or b >= 0 then sqrt(ab) = sqrt(a)sqrt(b) For any Real number x, at least one of 7-x >= 0 or 7+x >= 0, so we find: sqrt(49-x^2) = sqrt((7-x)(7+x)) = sqrt(7-x)sqrt(7+x)
How do you solve (x-1)^2=49? - Socratic 11 Jul 2015 · I found: x_1=8 x_2=-6 Take the square root of both sides and get: x-1=+-sqrt(49)=+-7 rearranging to isolate x you get two solutions: x_1=1+7=8 x_2=1-7=-6
What is the squared root of 144/49? - Socratic 12 Nov 2015 · sqrt(144/49)=12/7 sqrt(144/49) Simplify. (sqrt 144)/(sqrt49) 144=12xx12, therefore sqrt 144=12. 49=7xx7, therefore sqrt 49=7 sqrt(144/49)=12/7
How do you simplify #(49/81) ^ (-1/2)#? - Socratic 3 May 2018 · 9/7 1/sqrt(49/81) =1/(7/9) = 9/7. 5750 views around the world You can reuse this answer
How do you evaluate #\sqrt { - 81} + \sqrt { 49}#? - Socratic 17 Mar 2017 · Every non-zero number #n# has two square roots #sqrt(n)# and #-sqrt(n)#. The #sqrt# symbol denotes the principal square root which is defined as follows: If #n > 0# then #sqrt(n)# is the positive square root. If #n < 0# then #sqrt(n) = i sqrt(-n)# where #i# is the imaginary unit. In our example: #sqrt(-81)+sqrt(49) = sqrt(-9^2)+sqrt(7^2)#
How do you simplify #sqrt(49x^2)#? - Socratic 9 May 2016 · sqrt(49x^2)=7x Given - sqrt(49x^2) You can rewrite it like this sqrt(49) xx sqrtx^2 Then sqrt(49x^2)=sqrt(49) xx sqrtx^2=7x