How to use the discriminant to find out how many real number roots an equation has for $\frac{4}{3} x^{2} - 2 x + \frac{3}{4} = 0$ $4/3x^2 - 2x + 3/4 = 0$ ?

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Answer 1 · 2015-05-16T08:37:11

$(\frac{4}{3}) x^{2} - 2 x + \frac{3}{4}$ $(4/3)x^2-2x+3/4$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$

with $a = \frac{4}{3}$ $a=4/3$ , $b = - 2$ $b=-2$ and $c = \frac{3}{4}$ $c=3/4$ .

The discriminant $Delta$ is given by the formula $b^2-4ac$

In our case:

$Delta = b^2-4ac$

$= (-2)^2 - 4*(4/3)*(3/4)$

$= 4 - 4 = 0$

If $Delta > 0$ then our equation would have 2 real roots.

If $Delta < 0$ then our equation would have 2 complex non-real roots.

In our case $Delta = 0$ implies that our equation has 1 repeated real root.

That root is

$x = (-b +- sqrt(Delta)) / (2a) = (-(-2)+-0)/(2*(4/3)) = 2/(2*(4/3)) = 1/(4/3) = 3/4$

How to use the discriminant to find out how many real number roots an equation has for 4/3x^2 - 2x + 3/4 = 043x2−2x+34=04/3x^2 - 2x + 3/4 = 0?