Question

How do I, Solve `y=2-3x and y=x^2-2` algebraically (using the discriminant to find the general answer) and then graphically. Verify your solution using substitution. "?

Answer 1

`x= -4 and x=1`

Explanation:

Equate to each other through `y`

`2-3x=y=x^2-2`

`2-3x=x^2-2`

`x^2+3x-4=0`

Consider standard form `y=ax^2+bx+c ->x=(-b+-sqrt(b^2-4ac))/(2a)`

Where in this case: `a=1; b=3 and c=-4`

`=>x=(-3+-sqrt(3^3-4(1)(-4)))/(2(1))`

`x=(-3+-sqrt(25))/2`

`x=-3/2+-5/2`

`x= -4 and x=1`

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`color(blue)("Check for "x=-4)`

`2-3xcolor(white)("dddd")=y=color(white)("d")x^2-2`
`2-3(-4)=y=(-4)^2-2`

`color(white)("dddd")14color(white)("ddd")=y=color(white)("ddd")14 larr" Confirmed"`

`color(blue)("Check for "x=1)`

`2-3xcolor(white)("d.")=y=x^2-2`
`2-3(1)=y=(1)^2-2`

`color(white)("dddd")-1color(white)("ddd")=y=color(white)("ddd")-1 larr" Confirmed"`
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