How do you determine the number of real solutions to $- 3 x^{2} + 4 x + 1 = 0$ $-3x^2+4x+1=0$ ?

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How do you determine the number of real solutions to $- 3 x^{2} + 4 x + 1 = 0$ $-3x^2+4x+1=0$ ?

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Answer 1 · 2014-10-30T06:13:28

$- 3 x^{2} + 4 x + 1 = 0$ $-3x^2+4x+1=0$
$\Rightarrow 3 x^{2} - 4 x - 1 = 0$ $rArr 3x^2-4x-1=0$
$a = 3, b = - 4, c = - 1$ $a=3, b=-4, c=-1$

$D = b^{2} - 4 a c$ $D=b^2-4ac$
$\Rightarrow D = {(- 4)}^{2} - 4 (3) (- 1)$ $rArr D= (-4)^2 - 4 (3)(-1)$
$\Rightarrow D = 16 + 12$ $rArr D= 16 + 12$
$\Rightarrow D = 28$ $rArr D = 28$

As Discriminant is greater than zero, there are two real solutions.

PS :
If $D < 0$ $D < 0$ , there are no real solutions
If $D = 0$ $D = 0$ , there is only one real solution

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How do you determine the number of real solutions to $- 3 x^{2} + 4 x + 1 = 0$ $-3x^2+4x+1=0$ ?

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