Question

Use the quadratic formula to find the solutions to the equation. x²-3x+1=0 A. 3± √104 B. 3+√52 C. 21 √24 N D. 5√103

Answer 1

The correct solution to the quadratic equation
`x² - 3x + 1 = 0 isA. 3 ± √104.`

Step-by-step explanation:

The quadratic formula is given by
`\( x = (-b \pm √(b^2 - 4ac))/(2a) \),` where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0. In this case, the equation is
`x² - 3x + 1 = 0, so a = 1, b = -3, and c = 1.`

Now, substituting these values into the quadratic formula:

`\[ x = (-(-3) \pm √((-3)^2 - 4(1)(1)))/(2(1)) \]`

Simplifying further:

`\[ x = (3 \pm √(9 - 4))/(2) \]\[ x = (3 \pm √(5))/(2) \]`

This gives us two solutions:

`\[ x_1 = (3 + √(5))/(2) \]`

`\[ x_2 = (3 - √(5))/(2) \]`

These solutions can be further simplified by multiplying the numerator and denominator of each fraction by the conjugate of the denominator:

`\[ x_1 = (3 + √(5))/(2) \cdot (√(5) + 1)/(√(5) + 1) = (3√(5) + 3 + 5 + √(5))/(2(√(5) + 1)) \]`

`\[ x_2 = (3 - √(5))/(2) \cdot (√(5) - 1)/(√(5) - 1) = (3√(5) - 3 + 5 - √(5))/(2(√(5) - 1)) \]`

Simplifying both expressions, we get:

`\[ x_1 = (3 + √(5))/(2) = 3 + (√(5))/(2) \]`

`\[ x_2 = (3 - √(5))/(2) = 3 - (√(5))/(2) \]`

Thus, the solutions are
`\(3 + (√(5))/(2)\) and \(3 - (√(5))/(2)\),`which can be expressed as
`3 ± √5/2.`