Answer :
To find the roots of the quadratic equation [tex]\( x^2 - x - 42 = 0 \)[/tex], you can use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this formula, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the equation [tex]\(ax^2 + bx + c = 0\)[/tex]. For the equation [tex]\( x^2 - x - 42 = 0 \)[/tex], we have:
[tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -42\)[/tex].
Let's begin by calculating the discriminant, [tex]\( \Delta \)[/tex], which is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-1)^2 - 4(1)(-42) \][/tex]
[tex]\[ \Delta = 1 + 168 \][/tex]
[tex]\[ \Delta = 169 \][/tex]
Next, use the quadratic formula to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{169}}{2(1)} \][/tex]
[tex]\[ x = \frac{1 \pm 13}{2} \][/tex]
This gives us two solutions:
1. When [tex]\( \pm \)[/tex] is [tex]\( + \)[/tex]:
[tex]\[ x = \frac{1 + 13}{2} \][/tex]
[tex]\[ x = \frac{14}{2} \][/tex]
[tex]\[ x = 7 \][/tex]
2. When [tex]\( \pm \)[/tex] is [tex]\( - \)[/tex]:
[tex]\[ x = \frac{1 - 13}{2} \][/tex]
[tex]\[ x = \frac{-12}{2} \][/tex]
[tex]\[ x = -6 \][/tex]
Thus, the roots of the equation [tex]\( x^2 - x - 42 = 0 \)[/tex] are:
[tex]\[ x = 7 \][/tex]
and
[tex]\[ x = -6 \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this formula, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the equation [tex]\(ax^2 + bx + c = 0\)[/tex]. For the equation [tex]\( x^2 - x - 42 = 0 \)[/tex], we have:
[tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -42\)[/tex].
Let's begin by calculating the discriminant, [tex]\( \Delta \)[/tex], which is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-1)^2 - 4(1)(-42) \][/tex]
[tex]\[ \Delta = 1 + 168 \][/tex]
[tex]\[ \Delta = 169 \][/tex]
Next, use the quadratic formula to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{169}}{2(1)} \][/tex]
[tex]\[ x = \frac{1 \pm 13}{2} \][/tex]
This gives us two solutions:
1. When [tex]\( \pm \)[/tex] is [tex]\( + \)[/tex]:
[tex]\[ x = \frac{1 + 13}{2} \][/tex]
[tex]\[ x = \frac{14}{2} \][/tex]
[tex]\[ x = 7 \][/tex]
2. When [tex]\( \pm \)[/tex] is [tex]\( - \)[/tex]:
[tex]\[ x = \frac{1 - 13}{2} \][/tex]
[tex]\[ x = \frac{-12}{2} \][/tex]
[tex]\[ x = -6 \][/tex]
Thus, the roots of the equation [tex]\( x^2 - x - 42 = 0 \)[/tex] are:
[tex]\[ x = 7 \][/tex]
and
[tex]\[ x = -6 \][/tex]