Question

What are the coordinates of the `x`-intercepts of the graph of `y=2x^2+6x-20`?

Answer 1

Intercepts are `(-5,0)` and `(2,0)`

Explanation:

For `x`-intercepts we have their ordinates as zero i.e. `y=0`. Hence we get `x`-intercepts by putting `y=0`.

So we have `2x^2+6x-20=0`

or `x^2+3x-10=0`

or `x^2+2xx3/2xx x+(3/2)^2-(3/2)^2-10=0`

or `(x+3/2)^2-9/4-10=0`

or `(x+3/2)^2-49/4=0`

or `(x+3/2)^2-(7/2)^2=0`

or `(x+3/2+7/2)(x+3/2-7/2)=0`

or `(x+5)(x-2)=0` i.e. `x=-5` or `x=2`

Hence, intercepts are `(-5,0)` and `(2,0)`

graph{2x^2+6x-20 [-10, 10, -30, 30]}

Answer 2

X-intercepts are -

`(2,0)`
`(-5, 0)`

Explanation:

Given -

`y=2x^2+6x-20`

To find the x-intercept, put `y = 0`

`2x^2+6x-20=0`

Solve for `x`

`2x^2+6x=20`

`2/2x^2+6/2x=20/2` [divide both sides by coefficient of `x^2`]
`x^2+3x=10`

`x^2+3x+9/4=10+9/4` [divide the coefficient of `x` by `2` and square it. Add the value to both sides]
`x^2+3x+9/4=(40+9)/4=49/4`

`(x+3/2)^2=49/4`
`x+3/2=+-7/2`

`x=7/2-3/2=4/2=2`

`(2,0)`

`x=-7/2-3/2=-10/2=-5`

`(-5, 0)`