we are given polynomial 2x^2 +4x +7.
we need to determine the number of roots of the given polynomial using the fundamental theorem of algebra.
a) for the given polynomial 2x^2+4x+7, we have degree =2.
therefore, according to the fundamental theorem of algebra, the polynomial can have at most 2 roots.
b) let us find the roots of the given polynomial.
2x^2+4x+7= 0
![\mathrm{for\: a\: quadratic\: equation\: of\: the\: form\: }ax^2+bx+c=0\mathrm{\: the\: solutions\: are\: }](/tex.php?f=\mathrm{for\: a\: quadratic\: equation\: of\: the\: form\: }ax^2+bx+c=0\mathrm{\: the\: solutions\: are\: })
![\quad x_{1,\: 2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](/tex.php?f=\quad x_{1,\: 2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a})
![\mathrm{for\: }\quad a=2,\: b=4,\: c=7: \quad x_{1,\: 2}=\frac{-4\pm \sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2}](/tex.php?f=\mathrm{for\: }\quad a=2,\: b=4,\: c=7: \quad x_{1,\: 2}=\frac{-4\pm \sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2})
![x=\frac{-4+\sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2}=\quad -1+i\frac{\sqrt{10}}{2}](/tex.php?f=x=\frac{-4+\sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2}=\quad -1+i\frac{\sqrt{10}}{2})
![x=\frac{-4-\sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2}=\quad -1-i\frac{\sqrt{10}}{2}](/tex.php?f=x=\frac{-4-\sqrt{4^2-4\cdot \: 2\cdot \: 7}}{2\cdot \: 2}=\quad -1-i\frac{\sqrt{10}}{2})
therefore, roots are
![x=-1+i\frac{\sqrt{10}}{2}, \: x=-1-i\frac{\sqrt{10}}{2}](/tex.php?f=x=-1+i\frac{\sqrt{10}}{2}, \: x=-1-i\frac{\sqrt{10}}{2})
.