Question

What kind of roots does the equation `x^2+3x-4=0` have?

Answer 1

`x^2+3x-4=0` has two roots which are rational.

Explanation:

In the quadratic formula for quadratic equation `ax^2+bx+c=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

the discriminant is the part in the radical sign

`b^2-4ac`

It provides a simple test for the nature and number of solutions because:

• the solutions can only be real if the value of `b^2-4ac` is a positive number. A negative value would lea to the square root of a negative number, and is a complex value.

• there are two solutions if `b^2-4ac` is not equal to zero, since one solution would involve adding `b^2-4ac` to -b, and the other would require we subtract it. So, a zero value for `b^2-4ac` means only one root.

• Further, if `b` is a rational number and `b^2-4ac` is square of a rational number, not zero, we have two roots, which are rational.

So, without working out the entire formula, we can test only this small part, and determine the nature and number of the roots (the solutions to the equation).

In the given case we have `x^2+3x-4=0` i.e. `a=1`, `b=3` and `c=-4` and hence

`b^2-4ac=3^2-4xx1xx(-4)=9+16=25`,

which is square of a rational number, not zero, and hence

`x^2+3x-4=0` has two roots which are rational.