another math question on quadratic equations?

another math question on quadratic equations? Topic: another math question on quadratic equations?
June 19, 2019 / By Sloane
Question: Ok I'm lost again I understand how to factor the polynomial if the equation has numbers in the middle and the end such as x^2+8x+15=0 but what if the number is in the beginning such as 6x^2-x-2=0 or if there are values for all three such as 2x^2+13x+15=0 I am not looking for answers for homework but I need to understand how to work the problem. If you can explain it would help. Thanks in advance. i am not sure if you work it the same way as the other equation? I don't need the answer's just how to solve
Best Answer

Best Answers: another math question on quadratic equations?

Pollie Pollie | 10 days ago
Use the quadratic formula; -b± sqrt(b^2 -4(a)(c) all divided by 2(a) The formula in easier to understand terms is on wikipedia or you can do a google search of quadratic formula. Basically, quadratic equations have 3 terms, referred to as: a, b, c these letters are from the standard form of a quadratic equation: ax^2+bx+c = 0 the numbers in your problem correspond to a, b, and c - they are the coefficients of the terms in the equation so, a = 6, b = 1, c = -2 plug them into the formula and you will get your answer(s)
👍 288 | 👎 10
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Pollie Originally Answered: Quadratic equations and inequalities?
Start by drawing the graph of the equation. You should end up with a picture that looks like a "U", with two x-axis intercepts. It's important to note that the points that the graph intercepts the x-axis are points where y = 0. The points where the graph are above the x-axis are where y is greater than 0. The points below the x-axis are where y is less than 0. So, you are looking for 2x^2 - 5x - 3 >= 0. This means find all the values of x where y is greater or equal to zero. Those are all the points where the graph is above the x-axis. If you solve your function for its roots, you should get x = -0.5 and 3. On the graph you can see all the values of x <= -0.5 and x >= 3 will have y values greater than or equal to 0. (You can verify this by plugging any values in those ranges into the original function.) You know that it is x <= -0.5 because that means x points *to the left* of -0.5. And it is x >= 3.0 because that means x points *to the right* of x=3.0. Also, in both cases you know that it is >= and <= (not just > and <) because the original question includes =. So, the answer is x<=-0.5 and x>=3.0.

Matty Matty
you can learn how to complete the square which is a very complicated process and very annoying, or you can learn the quadratic formula, which looks complicated but is really very simple. For any quadratic ax^2+bx+c=0, x=0 at (-b plus or minus the square root of (b^2-4ac))/(2a) It's hard to type it neatly, but if you Google quadratic formula, you can find a nice and neat version. Remember that since a factor (x-a) means that the quadratic has a zero at a so just change your answer to a negative and subtract it from x. If you want to factor them neatly, then you have to complete the square or use trial and error. There are lots of sites on the internet that will show you the completing the square technique, but it's a really hard way to do the quadratic formula so I would just use that.
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Lashawn Lashawn
6x^2-x-2=0 you need to factor the numbers of the first and the third term, for example in this case its 6 and (-2). 6 = 2,3 or -2,-3 or 1,6 or -1,-6 -2 = 1,-2 or -1,2 from these combination by trial and error make a pair that will produce the middle term in this case its -1, for example you chose 2,3 for 6 -1,2 for -2 the middle term will be (2)(2)+(3)(-1) = 4 +(-3) = 1 this is not the correct combination 2,3 for 6 1,-2 for -2 the middle term will be (2)(-2)+(3)(1) = (-4) + 3 = -1 then this is it therefore the factors will be 2x + 1 3x -2 I hope this could help.
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Jessa Jessa
Same basic idea The first terms of each binomial factor have to multiply together to give the first term of the quadratic The last terms have to be selected to give a) the last term, when multiplied together b) the middle term under the OI operation in FOIL For 2x^2+13x+15, this is (2x+3)(x+5) And doing it is practice,practice,practice. A little later on you will learn the Quadratic Formula, which will give the roots (i.e.x= -3/2 and x= -5) for ANY quadratic, and hence its factors, (2x+3) and (x+5)
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Florinda Florinda
you use the tv table. ex: 6x^2+7x-2=0 you put possible factors under the first number in this case 6, and possible factors for the last number, 2. 6x^2+7x-2=0 ^______^ 2 - _- 1 then you make an x in between the number like so 3 - ' - 2 and you follow each line and multiply them: 2x2=4 3x1=3 then You take those two products and when you add then up they have to equal the middle number in this case 7. and i works, 3+4=7 :) if it didn't work out then you would have to try other possible factors under each number.
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Dalya Dalya
they work the same way. in x^2 + 8x +15 = 0 the implicit number on x^2 is 1 If their is a number other than 1 in front of the X^2 factor it in the same way 6x^2 -x -2 = 0 6 has factors of 1,.2,3,and 6 -2 has factors of -1,1,2,-2 in this case (3x - 2)(2x + 1) = 0 2x^2 + 13x +15 2 factors 1,2,-1,-1 15 factors1,3,5,15 (2x+3)(x+5) =0 x = (-1.5, -5)
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Blossom Blossom
first factor out a number if you can (but can't in your given examples otherwise, it's tougher, but you go through the same process, with (2x+?)(x-?) for example it takes longer, but the whole 2x, x set up is what you have to do
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Aisling Aisling
If ax^2 + bx +c is the equation,it can be written as a(x^2 +b/a x +c/a)= =a{x-[(-b/a) +(b^2/a^2 -4c/a)^1/2]/2} X {x-[(-b/a) - (b^2/a^2 -4c/a)^1/2]/2}
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Aisling Originally Answered: Math Homework.Quadratic Functions?
Don't be put off by the use of the term f(x). It is just a way of defining some other variable (often referred to as "y") in terms of x. First of all, the line of symmetry. Think about where this is found. We have a quadratic expression so the line of symmetry will be a vertical line, which will cross the curve at it's lowest point. What we do know is that at the lowest point of the curve, the gradient is zero. Or, if you prefer, the tangent to the curve is a horizontal line. The gradient of the curve at any point will be established by differentiating the quadratic equation. Differentiating 3x²-12x-63 using the basic "multiply by the power and substract one from the power" rule gives us: dy/dx f(x) = 6x-12 So at the tangent to the lowest point, which is perpendicular to the line of symmetry: 6x-12 = 0 or x = 2. This is our answer, since it is also the equation of a vertical line. For the co-ordinates of the vertex itself, we simply put the value of x back in to our original equation. So y = 3*(2²) - 12*2 - 63 y = 12 - 24 - 63 y = -75. So the vertex is at (2,-75). The y-intercept is fairly simple. At the y-intercept, the point where the curve crosses the y-axis, x=0. 3x² is 0 and so is 12x so the y value at the y-intercept is the constant in the equation, which is -63. The y-intercept is therefore (0,-63). The x-intercepts are a little more tricky. We need to factorise the equation first. First of all take out the common factor of 3, so: f(x) = 3 (x² - 4x - 21) Now further factorise the expression in the brackets. We need two values with a sum of -4 and a product of -21. It's generally easier to look at pairs which give the required product first. We could use 1 and -21, 3 and -7, 7 and -3 or 21 and -1. Only 3 and -7 however give us a sum of -4, the co-efficient of x. So the expression in the brackets factorises to give (x+3)(x-7). Our function is therefore: f(x) = 3(x+3)(x-7) At each x-intercept, where the curve crosses the x-axis, y=0. So if the function of x equals 0, either (x+3) is zero or (x-7) is zero. x is therefore either -3 or 7, and our x-intercepts are therefore (-3, 0) and (7,0).

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