Question

Quadratic formula decimal answer solve the equation to the nearest tenth

Answer 1

First, recall that given the equation

`ax^2+b+c=0`

The solutions are given by

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

where the sign in the middle means that we get one root by taking a plus sign and we get the other root by taking a minus sign.

In our case, we have a=1, b=-6 and c=-41. So the solutions of this equation are given by

`x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(-41)}}{2}=\frac{6\pm\sqrt[]{36+164}}{2}=\frac{6\pm\sqrt[]{200}}{2}`

Note that

`\sqrt[]{200}=\sqrt[]{100\cdot2}=\sqrt[]{100}\cdot\sqrt[]{2}=10\sqrt[]{2}`

Then

`x=\frac{6\pm10\sqrt[]{2}}{2}=3\pm5\sqrt[]{2}`

taking sqrt(2) as 1.4142, we get

`x=3+5\cdot\sqrt[]{2}=10.071\approx10`

and

`x=3-5\sqrt[]{2}=-4.071\approx-4`