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Vinil7 [7]
3 years ago
10

What is the solution set of the following equation?

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0

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}
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