Question

If x1 and x2 are solutions to 3x^2+15x+9=0 quadratic equation, then

x1+x2=

Answer 1

`x_1 + x_2 = \boxed{-5}`

Explanation:

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}`

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\implies x_(1,\:2)=(-b)/(2a)\pm\frac{\sqrt{b^(2)-4ac}}{2a}`

Let's have

`x_(1)=(-b)/(2a)+\frac{\sqrt{b^(2)-4ac}}{2a}`\

`x_(2)=(-b)/(2a)-\frac{\sqrt{b^(2)-4ac}}{2a}`

`x_1 + x_2 =(-b)/(2a)+\frac{\sqrt{b^(2)-4ac}}{2a} + (-b)/(2a)-\frac{\sqrt{b^(2)-4ac}}{2a}\\`

`\frac{\sqrt{b^(2)-4ac}}{2a} - \frac{\sqrt{b^(2)-4ac}}{2a} = 0\\\\`

Leaving us with

`(-b)/(2a) + (-b)/(2a)\\\\\implies 2 \cdot \left((-b)/(2a)\right)\\\\\implies -(b)/(a)\\\\`

The given equation is

`3x^2+15x+9=0`

Here a = 3, b = 15

So

`-(b)/(a) = --(15)/(3) = -5\\`

Answer:

`x_1 + x_2 = \boxed{-5}`