Best solutions to - "Elementary solutions of the equation of a quadratic..."

High School
Algebra
Algebra I
Quadratic function and equation

Jazmyn Holden

Answered question

2022-04-02

Elementary solutions of the equation of a quadratic formula
We know that when

A = 0

, the quadratic equation

A x^{2} + B x + C = 0

has one solution,

- \frac{C}{B}

Answer & Explanation

Jazlyn Mitchell

Beginner2022-04-03Added 14 answers

The solution $- \frac{C}{B}$ is the limit at 0 of one of those two solutions. Suppose that $B \geq 0$ . Then $\sqrt{B^{2}} = B$ and
$\begin{aligned} lim_{A \to 0} \frac{- B + \sqrt{B^{2} - 4 A C}}{2 A} & = \frac{1}{2} lim_{A \to 0} \frac{\sqrt{B^{2} - 4 A C} - B}{A} \\ = \frac{1}{2} {\frac{d}{d A} \sqrt{B^{2} - 4 A C} |}_{A = 0} \\ = - \frac{C}{\sqrt{B^{2}}} \\ = - \frac{C}{B} . \end{aligned}$
The case in which $B \leq 0$ is similar.

Mikaela Winters

Beginner2022-04-04Added 14 answers

You can use the conjugate quantity to get an alternative expression of the roots:
Suppose that

B > 0

. Then:

\frac{- B + \sqrt{B^{2} - 4 A C}}{2 A} = \frac{(- B + \sqrt{B^{2} - 4 A C}) (- B - \sqrt{B^{2} - 4 A C})}{2 A (- B - \sqrt{B^{2} - 4 A C})} = \frac{{(- B)}^{2} - {(\sqrt{B^{2} - 4 A C})}^{2}}{2 A (- B - \sqrt{B^{2} - 4 A C})} = \frac{2 C}{- B - \sqrt{B^{2} - 4 A C}}

A = 0

, the denominator is

- B - \sqrt{B^{2}} = - B - | B | = - 2 B

.
Therefore the root is

- \frac{C}{B}

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