Question

How do use the discriminant to find all values of b for which the equation `x^2-bx+4=0` has one real root?

Answer 1

`b=+-4`

Explanation:

The discriminant of a quadratic is `b^2-4ac`.

In this case, `a=1, b=?, c=-4`.

Equal roots means `b^2-4ac=0`

`b^2-4(1*4=b^2-16=0`

`b^2=16`

`b=+-4`

Answer 2

`b=+-4.`

Explanation:

For the Qudr. Eqn. `ax^2+bx+c=0,` to have only One Root, or, to

have Two Identical Roots,

`"The Dicriminant "Delta=b^2-4ac=0.`

In our case, `(-b)^2-4(1)(4)=b^2-16=0 rArr b=+-4.`

Incidentally, if

`b=4," the root is, "2; &, if b=-4," the root is "-2.`

Answer 3

In `y=ax^2+bx+c` we have:

`b=2sqrt(ac)=+-4`

Explanation:

Given the standard form `y=ax^2+bx+c`

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

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

For there to only be 1 real root we need the form

`x=(-b)/(2a)+-0`

This means that

`sqrt(b^2-4ac)/(2a)=0`

Multiply both sides by `2a`

`sqrt(b^2-4ac)=0`

square root both sides

`b^2-4ac=0`

`b=sqrt(4ac)=2sqrt(ac)`

But `a=1 and c=4` giving

`b=2sqrt(1xx4)=+-4`