Question

How do you find the solution to the quadratic equation `x^2 - 4x -3 = 0`?

Answer 1

`x=2+-sqrt7`

Explanation:

`"there are no whole numbers which multiply to - 3"`
`"and sum to - 4"`

`"we can solve using the method of "color(blue)"completing the square"`

`"the coefficient of the "x^2" term is 1"`

`• " add subtract "(1/2"coefficient of the x-term")^2" to"`
`x^2-4x`

`rArrx^2+2(-2)xcolor(red)(+4)color(red)(-4)-3=0`

`rArr(x-2)^2-7=0`

`rArr(x-2)^2=7`

`color(blue)"take the square root of both sides"`

`rArrx-2=+-sqrt7larrcolor(blue)"note plus or minus"`

`rArrx=2+-sqrt7larrcolor(red)"exact solutions"`

Answer 2

x = `2+- sqrt(7)`

Explanation:

Apply quadratic formula for this equation instead of trying to factor it out.
1/ `((-b+-sqrt((b)^2-4(a)(c)))/(2(a)))`

2/ `((-(-4)+-sqrt((-4)^2-4(1)(-3)))/(2(1)))`

3/ `((4+-sqrt(16+12))/(2))`

4/ `((4+-2sqrt(7))/(2))` ( 2 cancel out)

5/ x = `2+-sqrt(7)`

Answer 3

`x=2+sqrt7 or x=2-sqrt7`

Explanation:

Here,

`x^2-4x-3=0`

`=>x^2-4x+4-7=0`

`=>(x-2)^2=7=(sqrt7)^2`

`=>x-2=+-sqrt7`

`=>x=2+-sqrt7`

OR

Comparing with quadratic equation,

`ax^2+bx+c=0=>a=1,b=-4,c=-3`

`triangle=b^2-4ac=(-4)^2-4(1)(-3)`

`=>triangle=16+12=28=4xx7`

`sqrt(triangle)=2sqrt7`

So,

`x=(-b+-sqrt(triangle))/(2a)`

`x=(4+-2sqrt7)/(2(1))`

`x=2+-sqrt7`