Answer:
The two root of the given quadratic equation [tex]x^2-4x-9=29[/tex] is 8.48 and -4.48 .
Step-by-step explanation:
Consider, the given Quadratic equation, [tex]x^2-4x-9=29[/tex]
Thus on simplification, we get,
[tex]x^2-4x-9-29=0 \Rightarrow x^2-4x-38=0[/tex]
We can solve using quadratic formula,
For a given quadratic equation [tex]ax^2+bx+c=0[/tex] we can find roots using,
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] ...........(1)
Where, [tex]\sqrt{b^2-4ac}[/tex] is the discriminant.
Here, a = 1 , b = -4 , c = -38
Substitute in (1) , we get,
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot 1 \cdot (-38)}}{2 \cdot 1}[/tex]
[tex]\Rightarrow x=\frac{4\pm\sqrt{168}}{2}[/tex]
[tex]\Rightarrow x_1=\frac{4+\sqrt{168}}{2} and \Rightarrow x_2=\frac{4-\sqrt{168}}{2}[/tex]
Also, [tex]\sqrt{168}=12.96[/tex](approx)
[tex]\Rightarrow x_1=\frac{16.96}{2} and \Rightarrow x_2=\frac{-8.96}{2}[/tex]
[tex]\Rightarrow x_1=8.48 and \Rightarrow x_2=-4.48[/tex]
Thus, the two root of the given quadratic equation [tex]x^2-4x-9=29[/tex] is 8.48 and -4.48 .
