Question

true or false10. is (1+i) is a root of a quadratic equation with real coefficients, the equation is y=x2+2x+2

Answer 1

Given

`y=x^2+2x+2`

a=1

b=2

c=2

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ x=\frac{-2\pm\sqrt[]{2^2-4*1*2}}{2*1} \end{gathered}`
`\begin{gathered} x=\frac{-2\pm\sqrt[]{4^{}-4*1*2}}{2*1} \\ \\ x=\frac{-2\pm\sqrt[]{4^{}-8}}{2} \\ \\ x=\frac{-2\pm\sqrt[]{-4}}{2} \\ \\ \text{The root is complex} \\ \frac{-2\pm2i_{}}{2} \end{gathered}`
`\begin{gathered} \text{Simplify} \\ (-2)/(2)\pm(2i)/(2) \\ \\ \text{simplify the fraction} \\ x=\text{ -1}\pm1i \\ \text{which becomes} \\ x=-1+1i \\ x=-1-1i \end{gathered}`

Therefore, (1+i) is not a root, the only roots are; x= -1 +i and x= -1-i

The final answer is FALSE