# Thread: Discriminants and Real Solutions.

1. ## Discriminants and Real Solutions.

Hello everyone, I'm in need of a bit of help.
Could someone please explain the process of finding the discriminants and the number of real solutions to the quadratic equation of this expression:

$\displaystyle 5x^2+7x+6=0$

2. Originally Posted by tony_rose88
Hello everyone, I'm in need of a bit of help.
Could someone please explain the process of finding the discriminants and the number of real solutions to the quadratic equation of this expression:

$\displaystyle 5x^2+7x+6=0$
b^2-4ac
=(7)^2-4(5)(6)
=-71

The discriminant is negative, therefore no real roots.

3. For the polynomial $\displaystyle ax^{2} + bx + c = 0$, the discriminant is given as: $\displaystyle b^{2} - 4ac$. There are 3 cases in the value the discriminant can have and each tell you something about your quadratic/parabola:

$\displaystyle b^{2} - 4ac > 0$ - The quadratic has two real roots (crosses x-axis twice)
$\displaystyle b^{2} - 4ac = 0$ - The quadratic has exactly one real root (crosses x-axis once)
$\displaystyle b^{2} - 4ac < 0$ - The quadratic has no real roots (never crosses x-axis)

4. Thank you so much, both of you.
I understand it now!

But lets pretend that the Discriminants IS a positive number, how would I know how many Real Solutions there are?

I have one more quick question totally unrelated while I have you here:

I'm terrible at Simplifying Square Roots.
Could you explain how to simplify this expression for me?

$\displaystyle \sqrt(3y^2w^6) \sqrt(6yw^5)$ (not sure how to do the square root symbol on here)
Assume that all variables represent positive real numbers.

Thanks again!

5. Just two. For polynomials in the form of $\displaystyle ax^{2} + bx + c = 0$, you will not be able to draw the parabola crossing the x-axis more than twice.

6. \sqrt{xy} yields $\displaystyle \sqrt{xy}$.

You can rewrite the square root as an exponent of one-half, i.e.:

$\displaystyle \sqrt{a} = a^{\frac{1}{2}}$

For your question, you can use your power rule to help simplify them after rewriting the square root as an exponent:

$\displaystyle \left(a^{b}\right)^{c} = a^{b \cdot c}$

Something that maybe of help ... In general: $\displaystyle a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}$

7. $\displaystyle \sqrt{3y^2w^6} \sqrt {6yw^5}$

$\displaystyle = \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}$

Recall that:

$\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

So we have:

$\displaystyle = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}$

8. Originally Posted by topher0805
$\displaystyle \sqrt{3y^2w^6} \sqrt {6yw^5}$

$\displaystyle = \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}$

Recall that:

$\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

So we have:

$\displaystyle = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}$
Okay, I think I'm starting to get it, but could you explain to me where each of the 'parts' come from on this expression:
$\displaystyle = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

Also,
Is there another way to write that without powers as fractions?

I can't use fractions in my answer.
All variables represent positive real numbers.

Should I just turn them into decimals?

$\displaystyle = 3\cdot \sqrt{2}\cdot y^{1.5}\cdot w^{5.5}$ ?

9. No, don't put them into decimals.

$\displaystyle = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$
The $\displaystyle \sqrt{3}\cdot \sqrt{2}$ terms came from $\displaystyle \sqrt{6}$ .

$\displaystyle \sqrt{6} = \sqrt{3\cdot 2} = \sqrt{3}\cdot \sqrt{2}$

The last 4 terms came from this rule of exponents:

$\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

Another of doing this without fractions as exponents is as follows:

$\displaystyle = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot y\cdot \sqrt {y}\cdot w^{3}\cdot \sqrt {w^5}$

10. Another way of writing the final step is:

$\displaystyle 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}$

$\displaystyle = 3\cdot \sqrt{2}\cdot \sqrt {y^3}\cdot \sqrt {w^{11}}$

11. Ah, okay. I see!

Thank you so much!