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there is no problem with this quadratic equation it's 2x square - 3x -2 = 0 I need an answer. where it says square there should be a little 2 at the top corner of the 2x to make it 2x square thanks If you can't factor it easily then use the quadratic equation: The two solutions are: 2 & -.5
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∙ 15y ago
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What is the factor of 3x2 2x 2 equals 0?
3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9
What are the 2 solutions for x if 3x2 plus 2x-8 equals 0?
3x2+2x-8=0 Delta = 22-4*3*-8= 100 So x = (-2 + sqrt(100))/(2*3) or x = (-2 - sqrt(100))/(2*3) x = 4/3 or x=-2
When factored what is the solution set 3x squared minus 19x minus 14 equals 0?
3x2 - 19x - 14 = 0 3x2 - 21x + 2x - 14 = 0 3x(x - 7) + 2(x - 7) = 0 (x - 7)(3x + 2) = 0 x - 7 = 0 or 3x + 2 = 0 x = 7 or x = -2/3
4x2 plus 2x equals?
Assuming that 4x2+2x = 0: 2x (2x+1) = 0 2x = 0 or 2x+1 = 0 x = 0 or -1/2
Which is the solution set for 3x2 - 4x - 15 equals 0?
3x2-4x-15 = 0 (3x+5)(x-3) = 0 x = -5/3 or x = 3
How do you solve 3x squared plus 2x equals 16?
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
What is 3x squared plus 2x minus 8 equals 0?
3x2 + 2x - 8 = 0 is a quadratic equation.
X2 plus 8x equals -3x2 plus 5?
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
Factorize completely 3x2-5x plus 2?
If you meant 3x2 - 5x + 2, here is your answer 3x2 - 5x +2 = 0 => 3x2 - 3x - 2x + 2 = 0 => 3x(x-1) - 2(x-1) = 0 => (3x - 2) (x-1) = 0 => x = 2/3 or, x = 1
3x² -2x -2 factorize?
3x2-2x-2 = (3x-3.645751311)(x+0.5485837704) when factored with the help of the quadratic equation formula
What is 3x squared minus 2x minus 5 equals 0?
3x2 - 2x - 5 = (3x - 5)(x + 1) = 0 gives a solution of x = 5/3 or -1. Was that what you wanted?
What is the factor of 3x2 2x 2 equals 0?
3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9
What are the 2 solutions for x if 3x2 plus 2x-8 equals 0?
3x2+2x-8=0 Delta = 22-4*3*-8= 100 So x = (-2 + sqrt(100))/(2*3) or x = (-2 - sqrt(100))/(2*3) x = 4/3 or x=-2
Given the rational function 2x-3x2 -5x-6 find the verticle and horizontal asymptotes?
f(x) = (2x - 3x2)/(-5x - 6)f(x) = -(3x2 - 2x)/-(5x + 6)f(x) = (3x2 - 2x)/(5x + 6)5x + 6 = 05x + 6 - 6 = 0 - 65x = -65x/5 = -6/5x = -6/5 (the vertical asymptote of f)Since the degree of the polynomial function in the numerator is greater than the degree of the polynomial function in the denominator, then the graph of f has no horizontal asymptote.
Three times the square of a certain positive number exceeds six times the number by nine Find the number?
The problem can be written as3x2 = 6x + 9x is the number we want to findSolution:x2 = 2x + 3x2 - 2x - 3 = 0(x-3)(x+1) = 0x = -1, 3
What is the fifth step in solving the eqution by completing the square 3x2 plus 14x plus 8 equals 0?
3x2+14x+8 = 0 x = -2/3 or x = -4
Solve -3x² -2x -5 equals 0?
Are you sure the equation is not: -3x2 - 2x + 5 = 0? Which is the same as 3x2 + 2x - 5 = 0 which factorises into (3x + 5)(x - 1) with solutions x = 1 and x = -5/3 = -1 2/3 ~= -1.67. As stated there are no real solutions to the equation, only complex solutions: x = -1/3 + i sqrt(14)/3 and x = -1/3 - i sqrt(14)/3
