Using discriminants

**j0nath0n3** · August 30th 2008, 01:02 PM

Use the discriminant to determine the number of real solutions of the quadratic equations.

60. (-5xsquared - 4x + 4)= 0

62. (xsquared - 4x + 4) = 0

64. (4/7xsquared -8x + 28) =0

66. (9 + 2.4x - 8.3xsquared)=0

i don't know what it means by discriminant?

**o_O** · August 30th 2008, 01:10 PM

Given the quadratic $ax^2 + bx + c = {\color{red}0}$ (the 0 is important), the discriminant refers to ${\color{blue}b^2 - 4ac}$ which comes from the quadratic equation: $x = \frac{-b \pm \sqrt{{\color{blue}b^2 - 4ac}}}{2a}$

Now, if:

$b^2 - 4ac > 0$ , then there are 2 real solutions to x.
$b^2 - 4ac = 0$ , then there is exactly 1 real solution to x.
$b^2 - 4ac < 0$ , then there are no real solutions to x.

See if you can apply this to your quadratics.

**j0nath0n3** · August 30th 2008, 01:32 PM

Originally Posted by o_O

Given the quadratic $ax^2 + bx + c = {\color{red}0}$ (the 0 is important), the discriminant refers to ${\color{blue}b^2 - 4ac}$ which comes from the quadratic equation: $x = \frac{-b \pm \sqrt{{\color{blue}b^2 - 4ac}}}{2a}$

Now, if:

$b^2 - 4ac > 0$ , then there are 2 real solutions to x.
$b^2 - 4ac = 0$ , then there is exactly 1 real solution to x.
$b^2 - 4ac < 0$ , then there are no real solutions to x.

See if you can apply this to your quadratics.

Thanks you rock.

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