Question

What is the solution set of the following equation?

Answer 1

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=2,\:x=\frac{1}{2}$

Step-by-step explanation:

Given

$\frac{4}{5}x^2=2x-\frac{4}{5}$

$\mathrm{Multiply\:both\:sides\:by\:}5$

$\frac{4}{5}x^2\cdot \:5=2x\cdot \:5-\frac{4}{5}\cdot \:5$

$\mathrm{Simplify}$

$4x^2=10x-4$

$\mathrm{Add\:}4\mathrm{\:to\:both\:sides}$

$4x^2+4=10x-4+4$

$4x^2+4=10x$

$\mathrm{Subtract\:}10x\mathrm{\:from\:both\:sides}$

$4x^2+4-10x=10x-10x$

$4x^2-10x+4=0$

$\mathrm{Quadratic\:Equation\:Formula:}$

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

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

$\mathrm{For\:}\quad a=4,\:b=-10,\:c=4:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}$

$x=\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}$

  $=\frac{10+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}$

  $=\frac{10+\sqrt{36}}{2\cdot \:4}$      ∵ $10+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}=10+\sqrt{36}$

  $=\frac{10+\sqrt{36}}{8}$

  $=\frac{10+6}{8}$

  $=\frac{16}{8}$

Similarly,

$=\frac{-\left(-10\right)-\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}:\quad \frac{1}{2}$

Thus,

       $\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

• $x=2,\:x=\frac{1}{2}$