Question

How do you write a quadratic equation in standard form with solutions -2 and 7?

Answer 1

`color(green)(x^2-5x-14=0)`

Explanation:

If `color(red)("(-2))` and `color(blue)(""(7))` are solutions to a quadratic equation `color(green)(f(x)=0)`

then `(x-color(red)(""(-2)))` and `(x-color(blue)(7))` are factors of `color(green)(f(x)`
and we can write
`color(white)("XXX")color(green)(f(x))=(x+2)(x-7)=x^2-5x-14`

Answer 2

`a(x^2-5x-14)=0, a!=0`

Explanation:

Let the reqd. quadr. eqn. be `p(x)=ax^2+bx+c=0, ane0`.

`x=-2" is a soln. "rArr p(-2)=0 rArr 4a-2b+c=0... ... (1)`

Similarly, `p(7)=0 rArr 49a+7b+c=0.............................(2)`

`(2)-(1) rArr 45a+9b=0 rArr b=-5a.............................(3)`

Sub.ing this `b` in `(1), 4a+10a+c=0 rArr c=-14a.........(4)`.

Using `(3) and (4)`, we get the reqd. quadr.

`p(x)=ax^2-5ax-14a=0, or, p(x)=a(x^2-5x-14)=0, a!=0`