Question

What is the solution set of the following equation?

4/5x2 = 2x - 4/5
{1, 2}
{1, 1/2}
(2,1/2)​

Answer 1

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

`x=2,\:x=(1)/(2)`

Explanation:

Given

`(4)/(5)x^2=2x-(4)/(5)`

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

`(4)/(5)x^2\cdot \:5=2x\cdot \:5-(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)=(-b\pm √(b^2-4ac))/(2a)`

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

`x=(-\left(-10\right)+√(\left(-10\right)^2-4\cdot \:4\cdot \:4))/(2\cdot \:4)`

`=(10+√(\left(-10\right)^2-4\cdot \:4\cdot \:4))/(2\cdot \:4)`

`=(10+√(36))/(2\cdot \:4)` ∵
`10+√(\left(-10\right)^2-4\cdot \:4\cdot \:4)=10+√(36)`

`=(10+√(36))/(8)`

`=(10+6)/(8)`

`=(16)/(8)`

Similarly,

`=(-\left(-10\right)-√(\left(-10\right)^2-4\cdot \:4\cdot \:4))/(2\cdot \:4):\quad (1)/(2)`

Thus,

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

• 
  `x=2,\:x=(1)/(2)`