How do you describe the nature of the roots of the equation $x^{2} - 4 x + 4 = 0$ $x^2-4x+4=0$ ?

Question

10719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-17T20:11:14

We can factor the expression on the right as:

$(x - 2) (x - 2) = 0$ $(x - 2)(x - 2) = 0$ or ${(x - 2)}^{2} = 0$ $(x - 2)^2 = 0$

Therefore the two roots for this quadratic are the same: $x = 2$ $x = 2$

We can also use the discriminate to show this same result:

For $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

The discriminate is the portion of the quadratic equation within the radical: $b^{2} - 4 a c$ $color(blue)(b)^2 - 4color(red)(a)color(green)(c)$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$1$ $color(red)(1)$ for $a$ $color(red)(a)$

$- 4$ $color(blue)(-4)$ for $b$ $color(blue)(b)$

$4$ $color(green)(4)$ for $c$ $color(green)(c)$

${(- 4)}^{2} - (4 \cdot 1 \cdot 4) \Rightarrow$ $color(blue)((-4))^2 - (4 * color(red)(1) * color(green)(4)) =>$

$16 - 16 \Rightarrow$ $16 - 16 =>$

$0$ $0$

Because the discriminate is $0$ $0$ you get just ONE solution.

How do you describe the nature of the roots of the equation x2−4x+4=0x^2-4x+4=0?