Solve the quadratic equation 2x² - 4x + 1 = 0

How many solutions are in the equation 3x² + 4x - 7 = 0

(2x-1)^2=0

2x-1=0
x=1/2

in the next equation, it is a degree two equation, it has two solutions
x=(-4+-sqrt(16+12*7)/6= 1 and 2/3

2x² - 4x + 1 = 0

use the formula
x = (4 ± √(16 - 4(2)(1) )/4
= 3 ± √8)/4
= ...... , simplify it a bit more, remember √8 = 2√2

for the 2nd,
remember that the b^2 - 4ac of the formula is called the
discriminant,
if b^2 - 4ac > 0 there will be 2 and distinct roots
if b^2 - 4ac = 0 there will be 1 real root, actually 2 equal roots
if b^2 - 4ac < 0 you will have 2 different complex roots, one being the conjugate of the other.

To solve the quadratic equation 2x² - 4x + 1 = 0, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 2, b = -4, and c = 1. Plugging in these values, we get:

x = (-(-4) ± √((-4)² - 4(2)(1))) / (2(2))
= (4 ± √(16 - 8)) / 4
= (4 ± √8) / 4

Simplifying further, we have:

x = (4 ± 2√2) / 4

Now we can split this equation into two solutions:

x₁ = (4 + 2√2) / 4
= 1 + √2/2

x₂ = (4 - 2√2) / 4
= 1 - √2/2

Therefore, the quadratic equation 2x² - 4x + 1 = 0 has two solutions:

x₁ = 1 + √2/2
x₂ = 1 - √2/2

Now, let's move on to the second question.

To find out the number of solutions in the equation 3x² + 4x - 7 = 0, we can again use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 3, b = 4, and c = -7. Plugging in these values, we get:

x = (-4 ± √((4)² - 4(3)(-7))) / (2(3))
= (-4 ± √(16 + 84)) / 6
= (-4 ± √100) / 6
= (-4 ± 10) / 6

Simplifying further, we have:

x₁ = (-4 + 10) / 6
= 6 / 6
= 1

x₂ = (-4 - 10) / 6
= -14 / 6
= -7/3

Therefore, the quadratic equation 3x² + 4x - 7 = 0 has two solutions:

x₁ = 1 (double root)
x₂ = -7/3