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ashiteru123

ashiteru123

29.03.2020 •

Mathematics

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

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Ответ:

hcoulter15

22.01.2022

Step-by-step explanation:

Given

|x^2-x+1|=3

$\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:=\:a,\:a0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a$

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

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

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

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

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

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

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

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

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

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

so

$\mathrm{Combine\:Solutions:}$

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

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

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Ответ:

destinyw329

27.04.2022

Step-by-step explanation:

Hshsh

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