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

Jul 6, 2017

$b = \pm 4$

#### Explanation:

The discriminant of a quadratic is ${b}^{2} - 4 a c$.

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

Equal roots means ${b}^{2} - 4 a c = 0$

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

${b}^{2} = 16$

$b = \pm 4$

Jul 6, 2017

$b = \pm 4.$

#### Explanation:

For the Qudr. Eqn. $a {x}^{2} + b x + c = 0 ,$ to have only One Root, or, to

have Two Identical Roots,

$\text{The Dicriminant } \Delta = {b}^{2} - 4 a c = 0.$

In our case, ${\left(- b\right)}^{2} - 4 \left(1\right) \left(4\right) = {b}^{2} - 16 = 0 \Rightarrow b = \pm 4.$

Incidentally, if

$b = 4 , \text{ the root is, "2; &, if b=-4," the root is } - 2.$

Jul 6, 2017

In $y = a {x}^{2} + b x + c$ we have:

$b = 2 \sqrt{a c} = \pm 4$

#### Explanation:

Given the standard form $y = a {x}^{2} + b x + c$

where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$x = \frac{- b}{2 a} \pm 0$

This means that

$\frac{\sqrt{{b}^{2} - 4 a c}}{2 a} = 0$

Multiply both sides by $2 a$

$\sqrt{{b}^{2} - 4 a c} = 0$

square root both sides

${b}^{2} - 4 a c = 0$

$b = \sqrt{4 a c} = 2 \sqrt{a c}$

But $a = 1 \mathmr{and} c = 4$ giving

$b = 2 \sqrt{1 \times 4} = \pm 4$