Question

Find the roots of the following equation x^2-20x+98=0

Answer 1

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

x² - 20x + 98 = 0

Step 02:

ax² + bx + c = 0

a = 1

b = -20

c = 98

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

x1:

$x1\text{ = }\frac{-(-20)+\sqrt[]{(-20)^2-4\cdot1\cdot98}}{2\cdot1}$$x1\text{ = }\frac{20+2\sqrt[]{2}}{2}=10+\sqrt[]{2}$

x2:

$x2=\frac{-(-20)-\sqrt[]{(-20)^2-4\cdot1\cdot98}}{2\cdot1}$$x2\text{ = }\frac{20-2\sqrt[]{2}}{2}=10-\sqrt[]{2}$

The solution is:

$\begin{gathered} x1\text{ = 10 + }\sqrt[]{2} \\ x2\text{ = 10 - }\sqrt[]{2} \end{gathered}$