Answer:
Ques 1: [tex]-1\pm2i[/tex]
Ques 2: [tex](3x+5)[/tex]
Step-by-step explanation:
The quadratic formula states that the roots of the equation, [tex]ax^2+bx+c = 0[/tex] are given by, [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
Ques 1: The quadratic equation is given by, [tex]3x^2+6x+15 = 0[/tex].
On comparing, a= 3, b= 6 and c= 15.
So, the roots of the equation are given by,
[tex]x=\frac{-6\pm \sqrt{6^2-4\times 3\times 15}}{2\times 3}[/tex]
i.e. [tex]x=\frac{-6\pm \sqrt{36-180}}{6}[/tex]
i.e. [tex]x=\frac{-6\pm \sqrt{-144}}{6}[/tex]
i.e. [tex]x=\frac{-6\pm 12i}{6}[/tex]
i.e. [tex]x=\frac{-6+12i}{6}[/tex] and i.e. [tex]x=\frac{-6-12i}{6}[/tex]
i.e. [tex]x=-1+2i[/tex] and i.e. [tex]x=-1-2i[/tex]
Thus, the solutions of the equation are [tex]-1\pm2i[/tex].
Ques 2: The quadratic equation is [tex]9x^2+21x+10 = 0[/tex].
On comparing, a= 9, b= 21 and c= 10.
So, the roots of the equation are given by,
[tex]x=\frac{-21\pm \sqrt{21^2-4\times 9\times 10}}{2\times 9}[/tex]
i.e. [tex]x=\frac{-21\pm \sqrt{441-360}}{18}[/tex]
i.e. [tex]x=\frac{-21\pm \sqrt{81}}{18}[/tex]
i.e. [tex]x=\frac{-21\pm 9}{18}[/tex]
i.e. [tex]x=\frac{-21+9}{18}[/tex] and i.e. [tex]x=\frac{-21-9}{18}[/tex]
i.e. [tex]x=\frac{-12}{18}[/tex] and i.e. [tex]x=\frac{-30}{18}[/tex]
i.e. [tex]x=\frac{-2}{3}[/tex] and i.e. [tex]x=\frac{-5}{3}[/tex]
That is, the factors are [tex](3x+2)[/tex] and [tex](3x+5)[/tex]
So, according to the options, [tex](3x+5)[/tex] is the correct option.
