Answer:
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=2,\:x=\frac{1}{2}[/tex]
Step-by-step explanation:
Given
[tex]\frac{4}{5}x^2=2x-\frac{4}{5}[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:}5[/tex]
[tex]\frac{4}{5}x^2\cdot \:5=2x\cdot \:5-\frac{4}{5}\cdot \:5[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]4x^2=10x-4[/tex]
[tex]\mathrm{Add\:}4\mathrm{\:to\:both\:sides}[/tex]
[tex]4x^2+4=10x-4+4[/tex]
[tex]4x^2+4=10x[/tex]
[tex]\mathrm{Subtract\:}10x\mathrm{\:from\:both\:sides}[/tex]
[tex]4x^2+4-10x=10x-10x[/tex]
[tex]4x^2-10x+4=0[/tex]
[tex]\mathrm{Quadratic\:Equation\:Formula:}[/tex]
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\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}[/tex]
[tex]x=\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}[/tex]
[tex]=\frac{10+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}[/tex]
[tex]=\frac{10+\sqrt{36}}{2\cdot \:4}[/tex] ∵ [tex]10+\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}=10+\sqrt{36}[/tex]
[tex]=\frac{10+\sqrt{36}}{8}[/tex]
[tex]=\frac{10+6}{8}[/tex]
[tex]=\frac{16}{8}[/tex]
Similarly,
[tex]=\frac{-\left(-10\right)-\sqrt{\left(-10\right)^2-4\cdot \:4\cdot \:4}}{2\cdot \:4}:\quad \frac{1}{2}[/tex]
Thus,
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
- [tex]x=2,\:x=\frac{1}{2}[/tex]
