The excluded values for the expression x^2 + 11x + 18/x^2 - 9 is/are: __________

Question

The excluded values for the expression x^2 + 11x + 18/x^2 - 9 is/are: ___________ x = 3 only x = -3 only x = 0 only x = 3, x = -3 x = -9, x = -2 Add, and simplify your answer as much as possible: 4/3x^2 + 2x - 1 + 2/3x^2 - 4x + 1 = ___________ 2/(3x + 1)(x - 1) 3/x(3x + 1) 2/(3x - 1)(x + 1) 3/x(3x - 1) 2/(x - 1)(x + 1) Solve for u, where u is a real number: squareroot 40 - 6u = u - 4 __________ u = 6 u = -6, 6 u = 4 u = 20/3 No solution Use the quadratic formula to solve for x: 2x^2 + 5x - 1 = 0 ________ x = 0 x = -5 plusorminus squareroot 33/4 x = -5 plusorminus squareroot 17/4 x = -5 plusorminus squareroot 2/4 x = -5 plusorminus 3 squareroot 2/4

Explanation / Answer

17. The denominator of the given expression is x2 -9 = (x+3)(x-3). Since division by 0 is not defined, the excluded values of x are those which make the denominator 0 i.e. -3 and 3. Option D is the correct answer.

18.   We have 3x2+2x-1 = 3x(x+1)-1(x+1)= (x+1)( 3x-1) and 3x2 -4x+1 = 3x(x-1)-1(x-1) = (x-1)(3x-1).

Therefore,4/(3x2+2x-1)+2/(3x2-4x+1)= 4/(x+1)( 3x-1)+2/(x-1)(3x-1)= [4(x-1)+2(x+1)]/(x-1)(x+1)( 3x-1) = (4x-4+2x+2)/ (x-1)(x+1)( 3x-1) = ( 6x-2)/ (x-1)(x+1)( 3x-1)= 2(3x-1)/(x-1)(x+1)( 3x-1) = 2/(x-1)(x+1).Option E is the correct answer.

19.    We have (40-6u) = u-4. On squaring both the sides, we get 40-6u = (u-4)2 = u2 -8u +16 or, u2 -8u +16 +6u – 40 = 0 or, u2 -2u -24 = 0 or, u2 -6u+4u -24= 0 or, u(u-6) +4(u-6) = 0 or, (u-6)(u+4) = 0. Therefore, u = -4 or u = 6.Since the sign of the square root in the given expression is +, u = 6 is the solution.Option A is the correct answer.

20. We have 2x2 +5x -1 = 0. On using the quadratic equation, we have x = [ -5 ±{52 -4*2*(-1)}]/2*2 = [-5± (25+8)]/4 = (-5± 33)/4. Option B is the correct answer.

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