# If the fundamental theorem of algebra is correct , then why i am bein taught discriminant less than 0 for quadratic equations mean no real roots ? (i m a 10th standard student who was taught about polynomials and quadratic equations)

Nov 8, 2017

see explanation

#### Explanation:

If the discriminant is less than zero, there are no real roots, meaning that the graph never crosses the x-axis.

Example:

${x}^{2} - 2 x + 3$

$\Delta = - {2}^{2} - 4 \cdot 3$
$\Delta = - 4 - 8 = - 12$
$- 12 < 0$

This can also be seen from the graph

graph{x^2-2x+3 [-10, 10, -5, 5]}

This is important because then you can stress the validity of claims that might come up in questions like this.

1. Prove that this function has no real roots.

2. The discriminant can be used to solve equations

Example:
The equation
(p – 1)x^2 + 4x + (p – 5) = 0, where p is a constant
has no real roots.

(a) Show that p satisfies p^2 – 6p + 1=0

(b) Hence find the set of possible values of p.

This why everything about the discriminant is important

http://qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2013/Exam-materials/6663_01_que_20150513.pdf

Nov 8, 2017

The fundamental theorem of algebra allows for the roots to be either real or complex.

#### Explanation:

The fundamental theorem of algebra states that a polynomial of degree $n$ will have $n$ roots but doesn't state if they're real or complex.

So it's entirely possible for an ${n}^{\text{th}}$ degree polynomial to have $n$ real roots, $n$ complex roots or $k$ real and $n - k$ complex roots.

Looking at the example in the other answer, the polynomial has degree $2$ and the polynomial also has $2$ complex roots (x=1±isqrt2), so we see the theorem does actually hold.

Nov 8, 2017

If the discriminant is $< 0$, the roots will not be real numbers.

#### Explanation:

The discriminant is the quantity under the square root symbol in the quadratic formula.

The standard form for a quadratic equation:

$y = a {x}^{2} + b x + c$

To solve for $x$, the $0$ is substituted for $y$. You can now set up the quadratic formula to solve for the roots (values of $x$ when $y = 0$).

$x = \frac{- b \pm \sqrt{\textcolor{red}{{b}^{2}} - \textcolor{red}{4 a c}}}{2 a}$, where:

$\textcolor{red}{{b}^{2}} - \textcolor{red}{4 a c}$ is the discriminant.

If $\textcolor{red}{{b}^{2}} - \textcolor{red}{4 a c}$ is less than zero, then there will be no roots that are real numbers, and the solutions for $x$ will be complex numbers which will include $i$ to indicate that they are imaginary.

Example

$y = 5 {x}^{2} + 3 x + 1$

Substitute $0$ for $y$.

$0 = 5 {x}^{2} + 3 x + 1$

$a = 5$, $b = 3$, $c = 1$

Plug the values for $a$, $b$, and $c$ into the quadratic formula.

$x = \frac{- 3 \pm \sqrt{\textcolor{red}{{3}^{2}} - \textcolor{red}{4 \cdot 5 \cdot 1}}}{2 \cdot 5}$

Simplify.

$x = \frac{- 3 \pm \sqrt{\textcolor{red}{9} - \textcolor{red}{20}}}{10}$

Simplify the square root.

$x = \frac{- 3 \pm \sqrt{\textcolor{red}{- 11}}}{10}$

Apply $\sqrt{- x} = i \sqrt{x}$, where $i$ represents an imaginary number.

$x = \frac{- 3 \pm i \sqrt{11}}{10}$

Solutions for $x$.

$x = \frac{- 3 + i \sqrt{11}}{10} ,$$\frac{- 3 - i \sqrt{11}}{10}$

So if the discriminant is negative, the roots will not be real numbers.