How do you find the discriminant and how many solutions does $2 j^{2} - 3 j = - 1$ $2j^2 - 3j = -1$ have?

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How do you find the discriminant and how many solutions does $2 j^{2} - 3 j = - 1$ $2j^2 - 3j = -1$ have?

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Answer 1 · 2015-05-10T18:26:25

discriminant $D = b^{2} - 4 a c$ $D= b^2 - 4ac$

we have: $2 j^{2} - 3 j + 1 = 0$ $2j^2 - 3j + 1 = 0$

here:
$a = 2$ $a =2$ , $b = - 3$ $b =-3$ , $c = 1$ $c = 1$
(coefficients of $j^{2}$ $j^2$ , $j$ $j$ and the constant term respectively)

finding $D$ $D$ :
$D = b^{2} - 4 a c = (- 3^{2}) - (4 \times 2 \times 1)$ $D= b^2 - 4ac = (-3^2) - (4 xx 2 xx 1)$
$D = 9 - 8 = 1$ $D= 9-8 = 1$

formula for roots :
$j = \frac{- b \pm \sqrt{D}}{2 a}$ $j = (-b +- sqrt D) / (2a)$
$j = \frac{3 \pm \sqrt{1}}{2 \times 2}$ $j = (3 +- sqrt 1) / (2 xx 2)$
$j = \frac{3 + 1}{4} = \frac{4}{4} = 1 and \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$ $j = (3 + 1) / 4 = 4/4 = 1 and (3 -1) /4 = 2/4 = 1/2$
$j$ $j$ has two solutions:
$j = 1$ $j = 1$ and $j = \frac{1}{2}$ $j = 1/2$

the roots are real and unequal.

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How do you find the discriminant and how many solutions does $2 j^{2} - 3 j = - 1$ $2j^2 - 3j = -1$ have?

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