Question

Question 6 of 10 Select the two values of x that are roots of this equation. X2 - 5x + 2 = 0 DA. X= 5-17 2 B. x = 5- 33 2 NE C. 5+ 33 2 D. x= 5+ 17 2

Answer 1

Given equation

`x^2-5x+2=0`

use the quadratic equation,

`x_{}=(-b\pm√(b^2-4ac))/(2a)`
`a=1,\: b=-5,\: c=2`
`x_{}=(-\left(-5\right)\pm√(\left(-5\right)^2-4\cdot\:1\cdot\:2))/(2\cdot\:1)`

consider the discrement,

`\begin{gathered} √(\left(-5\right)^2-4\cdot\:1\cdot\:2) \\ =√(5^2-4\cdot\:1\cdot\:2) \\ =√(5^2-8) \\ =√(25-8) \\ =√(17) \end{gathered}`

Then,

`\begin{gathered} x_{}=(-\left(-5\right)\pm√(17))/(2\cdot\:1) \\ x_1=(-\left(-5\right)+√(17))/(2\cdot\:1),\: x_2=(-\left(-5\right)-√(17))/(2\cdot\:1) \\ \end{gathered}`
`\begin{gathered} x=(-\left(-5\right)+√(17))/(2\cdot\:1) \\ =(5+√(17))/(2\cdot\:1) \\ x=(5+√(17))/(2) \end{gathered}`
`\begin{gathered} x=(-\left(-5\right)-√(17))/(2\cdot\:1) \\ =(5-√(17))/(2\cdot\:1) \\ =(5-√(17))/(2) \\ \end{gathered}`

Answer : The solutions are, option A and D

`\begin{gathered} x=(5+√(17))/(2),\: x=(5-√(17))/(2) \\ (or) \\ x=\frac{5\pm\sqrt[]{17}}{2} \end{gathered}`