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# Equation with absolute values

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1. My teacher in class only showed us how to solve equations with absolute values by drawing a graph for it and then gave us a homework with the question as follows:

"Use algebra to find the exact solutions of the equation |2x2 + 6x - 5| = 5 - 2x"

How would this be done algebraically?
2. (Original post by Rexx18)
My teacher in class only showed us how to solve equations with absolute values by drawing a graph for it and then gave us a homework with the question as follows:

"Use algebra to find the exact solutions of the equation |2x2 + 6x - 5| = 5 - 2x"

How would this be done algebraically?
Start by finding the exact values which make the quadratic zero
3. x1 = (-3 +√19) / 2 and x2 = (-3 -√19) / 2

What next?
4. Consider what absolute value means. When the quadratic is positive, the equation is 2x^2+6x-5 = 5-2x so 2x^2+8x-10=0 so x^2+4x-5=0, giving (x-1)(x+5)=0 so x is 1 or 5. Now, when is the original quadratic positive? Well, 2x^2+6x-5>0 gives x<your second answer or x>your first answer (of the ones you found as told by M14B). Now the second answer evaluates to about 0.68 so both 1 and 5 work. Now consider what happens when x is between the values you found. So that 2x^2+6x-5 is negative. Then the equation becomes -2x^2-6x+5=5-2x so 2x^2+4x=0 -> x=0. Now, this is between -3.68 and 0.68, so it works. Thus the solutions are x=0, 1, or 5.
5. Ah, I see, thank you very much
6. Though your third solution should be -5 and not 5, since |2x2 + 6x - 5| with x = 5 gives 75 and 15 with x = -5
7. (Original post by Rexx18)
My teacher in class only showed us how to solve equations with absolute values by drawing a graph for it and then gave us a homework with the question as follows:

"Use algebra to find the exact solutions of the equation |2x2 + 6x - 5| = 5 - 2x"

How would this be done algebraically?
By algebraically, I just assume it means the standard way. The way i go about these I write down two versions:

2x2 + 6x - 5 = 5 - 2x

and

-(2x2 + 6x - 5) = 5 - 2x

Then proceed to solve them individually.

For drawing the graphs, it's really useful to have your axis on a fixed scale. First find where the quadratic inside the modulus signs equals 0, and secondly find the minimum point of the this quadratic. The minimum point will be reflected in y=0 due to modulus. Now begin to sketch the parabola, making it "bounce off" the x-axis where the quadratic equals 0, and then go through the new "minimum" point. For the line, well I'm sure you know how to sketch a line.

Sketches are useful to see how many points of intersection there are and form inequalities, but it is in no way the most accurate in finding solutions.
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FIRST:

2x2 + 8x - 10 = 0
x2 + 4x - 5 = 0
(x + 5)(x - 1)=0 therefore x=-5 and x=1

SECOND:

-(2x2 + 6x - 5) = 5 - 2x
2x2 + 6x - 5 = 2x - 5
2x2 + 4x=0
x(x+2)=0 therefore x=0 and x=-2

Solutions: x = 1, 0, -2, -5
8. (Original post by HapaxOromenon3)
Consider what absolute value means. When the quadratic is positive, the equation is 2x^2+6x-5 = 5-2x so 2x^2+8x-10=0 so x^2+4x-5=0, giving (x-1)(x+5)=0 so x is 1 or 5. Now, when is the original quadratic positive? Well, 2x^2+6x-5>0 gives x<your second answer or x>your first answer (of the ones you found as told by M14B). Now the second answer evaluates to about 0.68 so both 1 and 5 work. Now consider what happens when x is between the values you found. So that 2x^2+6x-5 is negative. Then the equation becomes -2x^2-6x+5=5-2x so 2x^2+4x=0 -> x=0. Now, this is between -3.68 and 0.68, so it works. Thus the solutions are x=0, 1, or 5.
Well, for one it should be -5, and secondly, you missed out -2 as a solution.

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