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This is a quadratic expression in x. It cannot be solved because it is an expression: there is no equation to solve. Furthermore, the discriminant for the quadratic = (-10)2 - 4*3*(-5) = 100+60 = 160 which is not a square number. There are, therefore no rational factors.
While it would be possible to give irrational factors, factorising a quadratic into linear terms containing irrational numbers is rarely useful.
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What is the point of intersection of the parabolas of y equals 3x squared plus 10x plus 11 and y equals 2 -2x -x squared?
They intersect at the point of: (-3/2, 11/4)
What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?
Points of intersection work out as: (3, 4) and (-1, -2)
What is 6-3x equals 5x-10x plus 2?
6-3x = 5x-10x+2 4 = 8x-10x 4 = -2x x = -2
What is the point of contact that the parabolas of y equals 2 -2x -x squared and y equals 3x squared plus 10x plus 11 make with each other showing work?
If: y = 2 -2x -x^2 and y = 3x^2 +10x +11 Then: 3x^2 +10x +11 = 2 -2x -x^2 So it follows that: 4x^2 +12x +9 = 0 Using the quadratic equation formula: x = -3/2 and x also = -3/2 Therefore by substitution the point of contact is at: (-3/2, 11/4)
How do you solve 3x squared plus 10x- 8 equals 0?
3x2 + 10x - 8 = 0 3x2 + 12x - 2x - 8 = 0 3x(x + 4) - 2(x + 4) = 0 (3x - 2)(x + 4) = 0 So 3x - 2 = 0 or x + 4 = 0 so x = 2/3 or x = -4
