Respuesta :
3x^2 + 6x - 10 = 0
3(x^2 + 2x) - 10 = 0
3[ (x + 1)^2 - 1 ] - 10 = 0
3(x+1)^2 - 13 = 0
so the vertex is at (-1,-13)
the roots will be same distance from x = -1
that is a distance 1.08 --1 = 2.08
so other root is approximately -1 -2.08 = -3.08
the other intercept is at (-3.08,0)
3(x^2 + 2x) - 10 = 0
3[ (x + 1)^2 - 1 ] - 10 = 0
3(x+1)^2 - 13 = 0
so the vertex is at (-1,-13)
the roots will be same distance from x = -1
that is a distance 1.08 --1 = 2.08
so other root is approximately -1 -2.08 = -3.08
the other intercept is at (-3.08,0)
Answer:
Step-by-step explanation:
The equation is given as:
[tex]y=3x^2+6x-10[/tex]
has 2 x-intercepts which can be found out by putting y=0 or by just solving the given equation.
But it is given that one of the x-intercept is 1.08.
For the above given equation [tex]y=3x^2+6x-10[/tex], sum of the roots is:
[tex]\frac{-b}{a}=\frac{-6}{3}=-2[/tex].
Let the other x-intercept be p, then
[tex]p+1.08=-2[/tex]
⇒[tex]p=-2+1.08[/tex]
⇒[tex]p=-3.08[/tex]
therefore, the another x intercept is (-3.08,0).