Question

In the quadratic equation above, k and p are constants. What are the solutions for x?
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Answer 1

Answer:

$\textsf{D)} \quad x=\dfrac{r}{2} \pm \dfrac{\sqrt{k^2+r^2}}{2}$

Step-by-step explanation:

Given equation:

$x^2-rx=\dfrac{k^2}{4}$

Subtract k²/4 from both sides:

$\implies x^2-rx-\dfrac{k^2}{4}=\dfrac{k^2}{4}-\dfrac{k^2}{4}$

$\implies x^2-rx-\dfrac{k^2}{4}=0$

Solve using the quadratic formula.

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=1 \quad \quad b=-r \quad \quad c=-\dfrac{k^2}{4}$

Substitute the values into the quadratic formula:

$\implies x=\dfrac{-(-r) \pm \sqrt{(-r)^2-4(1)\left(-\dfrac{k^2}{4}\right)} }{2(1)}$

$\implies x=\dfrac{-(-r) \pm \sqrt{(-r)^2-4\left(-\dfrac{k^2}{4}\right)} }{2}$

Apply the rule  (-a)² = a² :

$\implies x=\dfrac{-(-r) \pm \sqrt{r^2-4\left(-\dfrac{k^2}{4}\right)} }{2}$

Apply the rule  -(-a) = a :

$\implies x=\dfrac{r \pm \sqrt{r^2+\dfrac{4k^2}{4}} }{2}$

$\implies x=\dfrac{r \pm \sqrt{r^2+k^2} }{2}$

$\implies x=\dfrac{r \pm \sqrt{k^2+r^2} }{2}$

$\implies x=\dfrac{r}{2} \pm \dfrac{\sqrt{k^2+r^2}}{2}$