Question

What are the solutions of 3x2 + 15 = –6x?

–1 ± i
–1 ± 2i
–2 ± i
–2 ± 2i

Which of the following is a factor of 9x2 + 21x + 10?

3x + 5
9x + 2
9x + 5
None of the above

Answer 1

Ques 1:
`-1\pm2i`

Ques 2:
`(3x+5)`

Explanation:

The quadratic formula states that the roots of the equation,
`ax^2+bx+c = 0` are given by,
`x=(-b\pm √(b^2-4ac))/(2a)`.

Ques 1: The quadratic equation is given by,
`3x^2+6x+15 = 0`.

On comparing, a= 3, b= 6 and c= 15.

So, the roots of the equation are given by,

`x=(-6\pm √(6^2-4* 3* 15))/(2* 3)`

i.e.
`x=(-6\pm √(36-180))/(6)`

i.e.
`x=(-6\pm √(-144))/(6)`

i.e.
`x=(-6\pm 12i)/(6)`

i.e.
`x=(-6+12i)/(6)` and i.e.
`x=(-6-12i)/(6)`

i.e.
`x=-1+2i` and i.e.
`x=-1-2i`

Thus, the solutions of the equation are
`-1\pm2i`.

Ques 2: The quadratic equation is
`9x^2+21x+10 = 0`.

On comparing, a= 9, b= 21 and c= 10.

So, the roots of the equation are given by,

`x=(-21\pm √(21^2-4* 9* 10))/(2* 9)`

i.e.
`x=(-21\pm √(441-360))/(18)`

i.e.
`x=(-21\pm √(81))/(18)`

i.e.
`x=(-21\pm 9)/(18)`

i.e.
`x=(-21+9)/(18)` and i.e.
`x=(-21-9)/(18)`

i.e.
`x=(-12)/(18)` and i.e.
`x=(-30)/(18)`

i.e.
`x=(-2)/(3)` and i.e.
`x=(-5)/(3)`

That is, the factors are
`(3x+2)` and
`(3x+5)`

So, according to the options,
`(3x+5)` is the correct option.