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

3 Answers
Jul 6, 2017

Answer:

#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#

Jul 6, 2017

Answer:

# 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.#

Jul 6, 2017

Answer:

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#