Question

Solve 5x 2 - 7x + 2 = 0 by completing the square. What are the solutions?

Answer 1

`x_1=1\\\\x_2=(2)/(5)`

Explanation:

- You must divide the equation by 5:

`x^(2)-(7)/(5)x+(2)/(5)=0`

- Add and subtract
`(((7)/(5))/(2))^2`:

`x^(2)-(7)/(5)x+((7)/(10))^2+(2)/(5)-((7)/(10))^2=0`

Therefore, you obtain:

`(x-(7)/(10))^2-0.09=0`

-add 0.09} to both sides:

`(x-(7)/(10))^2=0.09`

- Apply square root to both sides and solve for x:

`\sqrt{(x-(7)/(10))^2}=√(\0.09)\\x-(7)/(10)=√(0.09)\\\\x_1=(7)/(10)+√(0.09)=1\\\\x_2=-(7)/(10)-√(0.09)=(2)/(5)`

Answer 2

`x=(2)/(5)` or
`x=1`

Explanation:

The given quadratic equation is

`5x^2-7x+2=0`

Group the constant terms on the right hand side.

`5x^2-7x=0-2`

`5x^2-7x=-2`

Divide through by 5.

`x^2-(7)/(5)x=-(2)/(5)`

Add the square of half the coefficient of x., which is
`((1)/(2)*- (7)/(5))^2=(49)/(100)` to both sides of the equation.

`x^2-(7)/(5)x+(49)/(100)=-(2)/(5)+(49)/(100)`

The left hand side is now a perfect square.

`(x-(7)/(10))^2=(9)/(100)`

Take the square root of both sides;

`(x-(7)/(10))=\pm \sqrt{(9)/(100)}`

`x-(7)/(10)=\pm (3)/(10)`

`x=(7)/(10)\pm (3)/(10)`

`x=(7-3)/(10)` or
`x=(7+3)/(10)`

`x=(4)/(10)` or
`x=(10)/(10)`

`x=1` or
`x=(2)/(5)`