Question

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

Answer 1

For the given equation, the solutions are $x=2 \pm\sqrt{11} .$

Step-by-step 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

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

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

$\frac{4 \pm \sqrt{16+28}}{2} = \frac{4 \pm \sqrt{44}}{2}.$

$\frac{4 \pm \sqrt{44}}{2} = \frac{4 \pm 2\sqrt{11}}{2} = 2 \pm\sqrt{11} .$

So $x=2 \pm\sqrt{11} .$