Question

How do you solve the equation |x^2-x+1|=3

Answer 1

Explanation:

Given

`|x^2-x+1|=3`

`\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:=\:a,\:a>0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a`

`x^2-x+1=-3\quad \mathrm{or}\quad \:x^2-x+1=3`

solving

`x^2-x+1=-3`

`x^2-x+1+3=-3+3`

`x^2-x+4=0`

`\mathrm{Discriminant\:}\:x^2-x+4=0`

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:discriminant\:is\:}b^2-4ac`

`\mathrm{For\:}\quad a=1,\:b=-1,\:c=4\quad`

`b^2-4ac=\left(-1\right)^2-4\cdot \:\:1\cdot \:\:4`

= 1 - 16

= -15

`\mathrm{Discriminant\:\:cannot\:be\:negative\:for\:}x\in \mathbb{R}`

`\mathrm{The\:solution\:is}`

`\mathrm{No\:Solution\:for}\:x\in \mathbb{R}`

similarly solving

`x^2-x+1=3`

`x^2-x+1-3=3-3`

`x^2-x-2=0`

`\mathrm{Solve\:with\:the\:quadratic\: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=1,\:b=-1,\:c=-2:\quad x_(1,\:2)=(-\left(-1\right)\pm √(\left(-1\right)^2-4\cdot \:1\left(-2\right)))/(2\cdot \:1)`

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

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

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

`x=2,\:x=-1`

so

`\mathrm{Combine\:Solutions:}`

`\mathrm{No\:Solution}\quad \mathrm{or}\quad \left(x=-1\quad \mathrm{or}\quad \:x=2\right)`

`x=-1\quad \mathrm{or}\quad \:x=2`