Question 1134282: Can please somebody help me solve this with explanation for a 8th grader. Thank you.
x^2-4|x|+3=0
Found 3 solutions by Theo, MathTherapy, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x^2 - 4|x| + 3 = 0
x is in between the absolute value signs.
that means that:
if x is positive, then |x| = x.
if x is negative, then |x| = -x
for example, if x is euqal to 5, the |5| = 5.
if x is eual to -5, then |-5| = - (-5) = 5.
the result is that the value of x is always positive.
you have two situations.
if x is positive, then x^2 - 4|x| + 3 = 0 becomes x^2 - 4x + 3 = 0
if x is negative, then x^2 - 4(-x) + 3 = 0 which becomes x^2 + 4x + 3 = 0
the first equation is x^2 - 4x + 3 = 0
factor this to get (x - 3) * (x - 1) = 0
the second equation is (x + 3) * (x + 1) = 0
the roots of the equation are x = -3, x = -1, x = 1, x = 3.
those are the values of the equation when y = 0.
if you want to graph this equation, you would look for the following intervals.
x < -3
x > -3 and < -1
x > -1 and < 1
x > 1 and < 3
x > 3
you would then evaluate the equation in each of those intervals to see if it is greater than 0 or less than 0.
when x = -4, y = x^2 - 4|x| + 3 gets you y = 3 which is positive.
when x = -2, y = x^2 - 4|x| + 3 gets you y = -1 which is negative.
when x = 0, y = x^2 - 4|x| + 3 gets you y = 3 which is positive.
when x = 2, y = x^2 - 4|x| + 3 gets you y = -1 which is negative.
when x = 4, y = x^2 - 4|x| + 3 gets you y = 3 which is positive.
you can now draw a rough sketch of your graph.
i did the same but used graphing software found at desmos.com.
here's what the graph looks like with the zero points being shown.
here's what the graph looks like with the test points being shown.
basically you're solving this the same way you solve any other absolute value problem.
you have to divide the function into two parts.
the first part is when the expression in the absolute value sign is positive.
the second part is when the expression in the absolute value sign is negative.
then solve each part separately.
Answer by MathTherapy(10552)  (Show Source):
Answer by ikleyn(52781)  (Show Source):
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