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What are the roots of the equatio
x² - x -42=0

Sagot :

konrad509

[tex]x^2-x-42=0\\x^2+6x-7x-42=0\\x(x+6)-7(x+6)=0\\(x-7)(x+6)=0\\x=7 \vee x=-6[/tex]

beinggreat78

Answer:

The roots are 7 and -6.

Keys:

  • [tex]ax^2+bx+c=0[/tex]
  • [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
  • [tex]-(-a)^(^n^) = a^(^n^)[/tex]
  • [tex]\sqrt[n]{a}^n = a[/tex]
  • [tex]1^a = 1[/tex]

When you see a ± in a quadratic equation, you will have multiple solutions; more than one.

Step-by-step explanation:

solve the main expression

[tex]x^2-x-42=0\\x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-42\right)}}{2\cdot \:1}\\\rightarrow \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-42\right)}\\=\sqrt{\left(-1\right)^2+4\cdot \:1\cdot \:42}\\\rightarrow \left(-1\right)^2\\=1^2\\=1\\\rightarrow 4\cdot \:1\cdot \:42\\=168\\=\sqrt{1+168}\\=\sqrt{169}\\=\sqrt{13^2}\\=13\\x_{1,\:2}=\frac{-\left(-1\right)\pm \:13}{2\cdot \:1}[/tex]

solve for x₁

[tex]\frac{-\left(-1\right)+13}{2\cdot \:1}\\=\frac{1+13}{2\cdot \:1}\\=\frac{14}{2\cdot \:1}\\=\frac{14}{2}\\=7[/tex]

solve for x₂

[tex]\frac{-\left(-1\right)-13}{2\cdot \:1}\\=\frac{1-13}{2\cdot \:1}\\=\frac{-12}{2\cdot \:1}\\=\frac{-12}{2}\\=-\frac{12}{2}\\=-6[/tex]

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