Question

Solve for k:

Answer 1

$\text{The value of k is } k = 0 \text{ or } k = \frac{-18}{47}$

Solution:

Given equation is:

$47k^2 +18k = 0$

We can solve the above equation by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\text{Compare } 47k^2 + 18k = 0 \text{ with } ax^2 + bx + c = 0$

$\mathrm{For\:}\quad a=47,\:b=18,\:c=0:\quad k_{1,\:2}=\frac{-18\pm \sqrt{18^2-4\cdot \:47\cdot \:0}}{2\cdot \:47}$

$k=\frac{-18+\sqrt{18^2-4\cdot \:47\cdot \:0}}{2\cdot \:47}\\\\\text{Simplify the above expression }\\\\k = \frac{-18 \pm \sqrt{324-0}}{94}\\\\k = \frac{-18 \pm \sqrt{324}}{94}\\\\k = \frac{-18 \pm 18}{94}$

Thus we have two solutions:

$k = \frac{-18+18}{94} \text{ or } k = \frac{-18-18}{94}\\\\k = 0 \text{ or } k = \frac{-36}{94}\\\\k = 0 \text{ or } k = \frac{-18}{47}$

$k = 0 \text{ or } k = \frac{-18}{47}$

$\text{Thus the value of k is } k = 0 \text{ or } k = \frac{-18}{47}$