Answer :
x² + 10x + 29 = 0
x = -(10) ± √((10)² - 4(1)(29))
2(1)
x = -10 ± √(100 - 116)
2
x = -10 ± √(-16)
2
x = -10 ± 4i
2
x = -5 ± 2i
x = -5 + 2i or x = -5 - 2i
THe roots of the function is 5 ± 2i.
x = -(10) ± √((10)² - 4(1)(29))
2(1)
x = -10 ± √(100 - 116)
2
x = -10 ± √(-16)
2
x = -10 ± 4i
2
x = -5 ± 2i
x = -5 + 2i or x = -5 - 2i
THe roots of the function is 5 ± 2i.
Answer:
The roots of the given equation are -5 + 2i and -5 - 2i
Step-by-step explanation:
Since, roots of an equation are the points which satisfy the equation,
Or the points which are obtained after solving the equation. ( by putting zero on right side )
Here, the given quadratic equation,
[tex]x^2+10x+29=0[/tex]
If we have an equation, ax² +bx + c = 0
By quadratic formula,
[tex]x=\frac{-b\pm \sqr{b^2-4ac}}{2a}[/tex]
By comparing,
The solution of the given equation is,
[tex]x=\frac{-10\pm \sqrt{10^2-4\times 1\times 29}}{2\times 1}[/tex]
[tex]x=\frac{-10\pm \sqrt{100-116}}{2}[/tex]
[tex]x=\frac{-10\pm \sqrt{-16}}{2}[/tex]
[tex]x=\frac{-10\pm +4i}{2}[/tex]
[tex]\implies x = -5\pm 2i[/tex]
Thus, the roots of the given equation are -5 + 2i and -5 - 2i