Question

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

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Answer 1

`\textsf{D)} \quad x=(r)/(2) \pm (√(k^2+r^2))/(2)`

Explanation:

Given equation:

`x^2-rx=(k^2)/(4)`

Subtract k²/4 from both sides:

`\implies x^2-rx-(k^2)/(4)=(k^2)/(4)-(k^2)/(4)`

`\implies x^2-rx-(k^2)/(4)=0`

Solve using the quadratic formula.

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

Therefore:

`a=1 \quad \quad b=-r \quad \quad c=-(k^2)/(4)`

Substitute the values into the quadratic formula:

`\implies x=\frac{-(-r) \pm \sqrt{(-r)^2-4(1)\left(-(k^2)/(4)\right)} }{2(1)}`

`\implies x=\frac{-(-r) \pm \sqrt{(-r)^2-4\left(-(k^2)/(4)\right)} }{2}`

Apply the rule (-a)² = a² :

`\implies x=\frac{-(-r) \pm \sqrt{r^2-4\left(-(k^2)/(4)\right)} }{2}`

Apply the rule -(-a) = a :

`\implies x=\frac{r \pm \sqrt{r^2+(4k^2)/(4)} }{2}`

`\implies x=(r \pm √(r^2+k^2) )/(2)`

`\implies x=(r \pm √(k^2+r^2) )/(2)`

`\implies x=(r)/(2) \pm (√(k^2+r^2))/(2)`