Question

Use the quadratic formula to solve the equation.

4x^2 - 10x + 5 = 0
Enter your answers, in simplified radical form.

X=_____ or X=_____​

Answer 1

`\large\boxed{x=(5-\sqrt5)/(4),\ x=(5+\sqrt5)/(4)}`

Explanation:

`\text{The quadratic formula for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then the equation has no real solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then the equation has one solution:}\ x=(-b)/(2a)\\\\\text{if}\ b^2-4ac,\ ,\ \text{then the equation has two solutions:}\ x=(-b\pm√(b^2-4ac))/(2a)\\\\==========================================`

`\text{We have the equation:}\ 4x^2-10x+5=0\\\\a=4,\ b=-10,\ c=5\\\\b^2-4ac=(-10)^2-4(4)(5)=100-80=20>0\\\\x=(-(-10)\pm√(20))/(2(4))=(10\pm√(4\cdot5))/(8)=(10\pm\sqrt4\cdot\sqrt5)/(8)=(10\pm2\sqrt5)/(8)\\\\=(2(5\pm\sqrt5))/(8)=(5\pm\sqrt5)/(4)`

Answer 2

Note that
`+\vee-` stands for plus or minus.

For the quadratic equation of the form
`ax^2+bx+c=0` the solutions are
`x_(1,2)=(-b+\vee-√(b^2-4ac))/(2a)` that means that for
`a=4, b=-10, c=5\Longrightarrow x_(1,2)=(-(-10)+\vee-√((-10)^2-4\cdot4\cdot5))/(2\cdot4)` this simplifies to
`\boxed{x_1=(5+√(5))/(4)}, \boxed{x_2=(5-√(5))/(4)}`

Hope this helps.