Question

X²-3x+2=0. x²-4x+3=0. x²-5x+4=0

Answer 1

$x^2-3x+2=0,\\(D=9-8=1,)\\x=\frac{3\pm1}{2},\\x_1=2,\\x_2=1.$

$x^2-4x+3=0,\\(D=16-12=4,)\\x=\frac{4\pm2}{2} ,\\x_1=3,\\x_2=1.$

$x^2-5x+4=0,\\(D=25-16=9,)\\x=\frac{5\pm3}{2},\\ x_1=4,\\x_2=1.$

Answer 2

To solve these quadratic equations, we can use the quadratic formula, which is given by:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions are:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each of the equations:

1. x² - 3x + 2 = 0 a = 1, b = -3, c = 2

x = (-(-3) ± √((-3)^2 - 412)) / (2*1) x = (3 ± √(9 - 8)) / 2 x = (3 ± √1) / 2

There are two solutions: x₁ = (3 + 1) / 2 = 4 / 2 = 2 x₂ = (3 - 1) / 2 = 2 / 2 = 1

So, the solutions to the first equation are x = 2 and x = 1.

2. x² - 4x + 3 = 0 a = 1, b = -4, c = 3

x = (-(-4) ± √((-4)^2 - 413)) / (2*1) x = (4 ± √(16 - 12)) / 2 x = (4 ± √4) / 2

There are two solutions: x₁ = (4 + 2) / 2 = 6 / 2 = 3 x₂ = (4 - 2) / 2 = 2 / 2 = 1

So, the solutions to the second equation are x = 3 and x = 1.

3. x² - 5x + 4 = 0 a = 1, b = -5, c = 4

x = (-(-5) ± √((-5)^2 - 414)) / (2*1) x = (5 ± √(25 - 16)) / 2 x = (5 ± √9) / 2

There are two solutions: x₁ = (5 + 3) / 2 = 8 / 2 = 4 x₂ = (5 - 3) / 2 = 2 / 2 = 1

So, the solutions to the third equation are x = 4 and x = 1.