Question

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

Answer 1

Answer:

The roots are 7 and -6.

Keys:

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

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

Step-by-step explanation:

solve the main expression

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

solve for x₁

$\frac{-\left(-1\right)+13}{2\cdot \:1}\\=\frac{1+13}{2\cdot \:1}\\=\frac{14}{2\cdot \:1}\\=\frac{14}{2}\\=7$

solve for x₂

$\frac{-\left(-1\right)-13}{2\cdot \:1}\\=\frac{1-13}{2\cdot \:1}\\=\frac{-12}{2\cdot \:1}\\=\frac{-12}{2}\\=-\frac{12}{2}\\=-6$

Answer 2

$x^2-x-42=0\\x^2+6x-7x-42=0\\x(x+6)-7(x+6)=0\\(x-7)(x+6)=0\\x=7 \vee x=-6$