0
3x^2 = 44 divide both sides by 3; x^2 == 44/3 x = +,- square root of 44/3 write 44 = 4 x 11; x = +,- 2(square root of 11/3) multiply the square root of 11/3 by square root of 3/3; x = +,- (2/3)(square root of 33)
square root of 4x is 2 times square root of x, so answer is square root of x times 3 since it is 2 square roots of x plus one of them
No. The Square root of x is not the value of x. So it can not be simplified beyond: Root X + root 3x Yes. The square root of 3x equals the square root of 3 times the square root of x, so when you add another square root of x, you can factor out the square root of x, thereby simplifying the expression to the square root of x times the sum of one plus the square root of three.
4 x square root of 10 minus square root of 10 = 3 x square root of 10.
Let the coefficient by 'x' Hence its square root is x^(1/2) or x^(0.5) Then the square root again is [x^(1/2)]^(1/2) Third time over {[x^(1/2)]^(1/2)}^(1/2) Now the rules of indices are [x^(n)[^(m) = x^(nm) When terms are 'nested' , multiply together. Also x^(n) X x^(m) = x^(n+m) x^)n) / x^(m) = x^(n-m) However, the first rule (nesting) applies in this case, when you multiply the indices together/ Hence x^(1/2 X 1/2 X 1/2) = x^(1/8) , Which is the 8th root.!!!!!
300*square root of 3 + 300*square root of 6 = 1254.462165
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
5
3x^2 = 44 divide both sides by 3; x^2 == 44/3 x = +,- square root of 44/3 write 44 = 4 x 11; x = +,- 2(square root of 11/3) multiply the square root of 11/3 by square root of 3/3; x = +,- (2/3)(square root of 33)
square root of 4x is 2 times square root of x, so answer is square root of x times 3 since it is 2 square roots of x plus one of them
3/(4*square root(x)) ....Mukesh
No. The Square root of x is not the value of x. So it can not be simplified beyond: Root X + root 3x Yes. The square root of 3x equals the square root of 3 times the square root of x, so when you add another square root of x, you can factor out the square root of x, thereby simplifying the expression to the square root of x times the sum of one plus the square root of three.
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
4 x square root of 10 minus square root of 10 = 3 x square root of 10.
-2 and -3Check:(-2) + (-3) = -5(-2)(-3) = 6Thus -2 and -3 are not the required numbers. let's find them: x + y = -6xy = -5 y = -x -6x(-x - 6) = -5-x^2 - 6x = -5x^2 + 6x = 5x^2 + 6x + 9 = 5 + 9(x + 3)^2 = 14x + 3 = (+ & -)square root of 14x = -3 (+ & -)square root of 14x = -3 + square root of 14 or x = - 3 - square root of 14y = -x - 6y = 3 - square root of 14 - 6 or y = 3 + square root of 14 - 6y = -3 - square root of 14 or y = -3 + square root of 14Check:(-3 + square root of 14) + (-3 - square root of 14) = -6(-3 + square root of 14)(-3 - square root of 14) = -5 ?(-3)^2 - (square root of 14)^2 = -5 ?9 - 14 = -5Check also tow other numbers.
root 27 = root 9 x root 3 = 3 root 3. Dividing this by root 3 gives 3!
Let the coefficient by 'x' Hence its square root is x^(1/2) or x^(0.5) Then the square root again is [x^(1/2)]^(1/2) Third time over {[x^(1/2)]^(1/2)}^(1/2) Now the rules of indices are [x^(n)[^(m) = x^(nm) When terms are 'nested' , multiply together. Also x^(n) X x^(m) = x^(n+m) x^)n) / x^(m) = x^(n-m) However, the first rule (nesting) applies in this case, when you multiply the indices together/ Hence x^(1/2 X 1/2 X 1/2) = x^(1/8) , Which is the 8th root.!!!!!