Answer:
(a) [tex]x=-2,3[/tex] (b) [tex]x=\frac{3}{4} ,-\frac{11}{4}[/tex]
Step-by-step explanation:
[tex]y=x^2-x-6\\y=2x^2+x-6[/tex]
For the first equation, this can easily be factored
[tex]y=(x+2)(x-3)[/tex]
Now, set each of their factors equal to zero and solve
[tex]x+2=0\\x=-2\\\\x-3=0\\x=3[/tex]
Therefore, for a, the roots of [tex]y=x^2-x-6[/tex] are [tex]x=-2, 3[/tex]
For the second equation, it is able to be factored, however, its normally easier to use the quadratic equation to solve polynomials with an a term greater than 1
[tex]y=2x^2+x-6[/tex]
[tex]x_{1} =-b+\frac{\sqrt{b^2-4ac} }{2a}=-1+\frac{\sqrt{1-4(2)(-6)} }{4}=-1+\frac{\sqrt{1-(-48)} }{4}=-1+\frac{\sqrt{49} }{4} =-1+\frac{7}{4}=\frac{3}{4}[/tex]
[tex]x_{2}=-b-\frac{\sqrt{b^2-4ac}}{2a}=-1-\frac{\sqrt{1-4(2)(-6)} }{4} =-1-\frac{\sqrt{1-(-48)} }{4}=-1-\frac{\sqrt{49} }{4}=-1-\frac{7}{4} =-\frac{11}{4}[/tex]
Hope this helps! :)
