A quadratic equation is shown below:

9x2 - 16x + 60 = 0

Part A: Describe the solution(s) to the equation by just determining the discriminant. Show your work. (3 points)

Part B: Solve 4x2 + 8x - 5 = 0 by using an appropriate method. Show the steps of your work and explain why you chose the method used. (4 points)

Part C: Solve 2x2 -12x + 5 = 0 by using a method different from the one you used in Part B. Show the steps of your work. (3 points)

Question

A quadratic equation is shown below:

9x2 - 16x + 60 = 0

Part A: Describe the solution(s) to the equation by just determining the discriminant. Show your work. (3 points)

Part B: Solve 4x2 + 8x - 5 = 0 by using an appropriate method. Show the steps of your work and explain why you chose the method used. (4 points)

Part C: Solve 2x2 -12x + 5 = 0 by using a method different from the one you used in Part B. Show the steps of your work. (3 points)

anonymous · Answer

help me please :(

anonymous · Answer

I'm trying to remember how to solve this kind of problem. The discriminant is -b+4ac, right?

anonymous · Answer

i think

anonymous · Answer

The discriminant is b^2-4ac, so that would be (16^2)-4(9)(60), which equals -1,904. This would have two complex roots because it is less than 0.

anonymous · Answer

thats part a

anonymous · Answer

But does that all make sense?

anonymous · Answer

4x^2 + 8x - 5 = 0
For this one, I would use the quadratic formula, and honestly that's just because I don't even think this is factorable. 
To start it, I would use the formula$$x=-b \pm \sqrt{b^2-4ac}/2a$$
4 is a, 8 is b, -5 is c. 
This gives you -8 $$-8\pm \sqrt{64-4(4)(-5)}/8$$
This is equal to $$-8\pm12/8$$
This is 1/2 or -5/2.

whpalmer4 · Answer

B is certainly factorable.

anonymous · Answer

It's actually better to factor part B by using (2x + 5)(2x - 1)= 0, and use the quadratic formula for part C.

anonymous · Answer

Yeah, I'm realizing that now. I'm just not good with factoring at all, so it takes a while.

whpalmer4 · Answer

Quick factoring of $$4x^2+8x-5$$by grouping:
multiply leading and trailing coefficients to get 4*-5 = -20
find pair of factors of -20 that sum to 8: -2*10 = -20, -2+10 = 8
rewrite, splitting middle term with two factors:
$$4x^2-2x + 10x -5$$Now group into two groups$$(4x^2-2x) + (10x-5)$$Factor each group:$$2x(2x-1)+5(2x-1)$$Now factor out common factor in each group:$$(2x-1)(2x+5)$$

whpalmer4 · Answer

@nani0710   your statement of the quadratic formula is not written correctly — you need to put parentheses around the numerator if you are going to write it all on one line like that, because the order of evaluation of expressions will have you divide the square root by the denominator before adding the leading term of what you intended to be the numerator!

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$is the correct way of writing it, or if you must write it all on one line:
$$x = (-b\pm\sqrt{b^2-4ac})/(2a)$$

You got the right answers, so I know you understand how it works, but when explaining it to someone else who may not, it's vital that you write it properly!

anonymous · Answer

thanks

anonymous · Answer

I know, and I usually would, but I'm not a frequent user on this website and I don't use the different ways of writing out equations very often, so it was my mistake. I'll make sure I provide that distinction next time.

whpalmer4 · Answer

Well, just a strategically placed set or two of parentheses would be all that you need to make it unambiguous.  Anyhow, just trying to make it easier for anyone who doesn't already understand what we're trying to show them :-)

anonymous · Answer

You didnt even do part C -.-

whpalmer4 · Answer

@karaa1108 It has been at least a year since the problem was posted, and the original poster has not complained about how part C was not completed. Perhaps they found the instructions (which included directions on how to do part C: "use the quadratic formula for part C") sufficient. 

However, in the interests of completeness:

$$2x^2 -12x + 5 = 0$$This is a quadratic in the form $$ax^2 + bx + c=0$$with $a=2,\ b=-12,\ c=5$ and can be solved with the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$if one did not use the quadratic formula for part B.

It is not factorable.  Graphing would be tedious and only approximate the answer (which has an irrational number).  You could also solve it by completing the square.

$$2x^2 - 12x + 5 = 0$$$$2x^2 - 12x = -5$$$$2(x^2-6x) =-5$$$$x^2-6x =-\frac{5}{2}$$Completing the square:$$x^2-6x = x^2 -6x +(\frac{6}{2})^2 - (\frac{6}{2})^2 = (x-\frac{6}{2})^2 - 9=(x-3)^2 -9$$Substituting back:$$(x-3)^2 - 9 =-\frac{5}{2}$$You should be able to solve it from there even if you don't know how to complete the square.