Question

What are the roots of x^2 - 4x - 7 = 0

Answer 1

For the given equation, the solutions are
`x=2 \pm√(11) .`

Explanation:

Step 1:

For an equation of the form
`ax^(2) +bx+c=0` the solution is
`x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}`.

Here a is the coefficient of
`x^(2)`, b is the coefficient of x and c is the constant term.

Comparing
`x^(2) -4x-7=0` with
`ax^(2) +bx+c=0`, we get that a is 1, b is -4 and c is -7.

To get the solution, we substitute the values of a, b, and c in
`x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}`.

Step 2:

Substituting the values, we get

`x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{(-4)^(2)-4(1)(-7)}}{2(1)}.`

`\frac{-(-4) \pm \sqrt{(-4)^(2)-4(1)(-7)}}{2(1)}= (4 \pm √(16+28))/(2).`

`(4 \pm √(16+28))/(2) = (4 \pm √(44))/(2).`

`(4 \pm √(44))/(2) = (4 \pm 2√(11))/(2) = 2 \pm√(11) .`

So
`x=2 \pm√(11) .`