With the variable as x and the a, b, and c the numbers in the quadratic equation
ax^2 +bx +c = 0
we can find the value for x that will make the above equation true by using this formula:
x = \frac{-b\pm\sqrt{b^2 -4ac} }{2a}
Solve the equation below. Do you see the right side is 0? One side must be 0. Then we must decide a, b, and c. { a (multiplied by x^2) is 3 and the b (multiplied by x) is -1 and c (the “lone number” is -2}
3x^2 -x – 2 = 0\qquad where \qquad a=3,\qquad b=-1,\qquad c=-2
x = \frac{-b\pm\sqrt{b^2 -4ac} }{2a}\qquad becomes
x=\frac{ +1\pm\sqrt{(-1)^2 -4(3)(-2)}}{2(3)}
Do you see that b is -1 so that means -b is +1. Now look under the square root and recall Order of Operation rules. We will get 1 – 12(-2) or 1+24.
x= \frac{1\pm\sqrt{25}}{6}\Rightarrow x= \frac{6}{6}=1, or\Rightarrow x= \frac{-4}{6}= \frac{-2}{3}
Now test your solutions (answers for x). Will the 1 make the quadratic expression above =0? Will the -2/3 also make the above expression =0? They should!
3x^2 -x – 2 = 0\\
becomes\\
3(1)^2 -(1) – 2 = 0\\
3 -1-2 =0\\
YES,\qquad this\qquad is\qquad true.
3x^2 -x -2 \qquad becomes \\
3(-2/3)^2 -(-2/3)-2=0 \\ and \qquad this\qquad is\\3(4/9)+2/3 -2 =0\\ which \qquad is \\ 4/3+2/3 -6/3=0\\ Is \qquad this \qquad true?\\6/3-6/3 =0, \qquad YES.
