Question

determine the nature of the roots of the following quadratic equation x²-4x+4=0

Answer 1

We can factor the expression on the right as:

(x−2)(x−2)=0 or (x−2)2=0

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

Explanation:

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

The quadratic formula states:

For ax2+bx+c=0, the values of x which are the solutions to the equation are given by:

x=−b±√b2−4ac2a

The discriminate is the portion of the quadratic equation within the radical: b2−4ac

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 for a

−4 for b

4 for c

(−4)2−(4⋅1⋅4)⇒

16−16⇒

0

Because the discriminate is 0 you get just ONE solution.



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Answer 2

Here is the answer

Given Equation =>
$\bf{ {x }^{2} - 4x + 4 = 0}$

Comparing the given Equation with ax^2 + bx + 0.
We get

a = 1
b = 4
c = 4

$\bf{\triangle = {b}^{2} - 4ac}$

$\bf{= (4) {}^{2} - 4(1) \times (4)}$

$\bf{= 16 - 16 = 0}$

Thus,
$\bf{\triangle = 0}$

The roots are real and equal.

Thanks!!