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The roots of the function [tex]f(x)=x^2-2x-3[/tex] are shown. What is the missing number?

[tex]x = -1[/tex] and [tex]x = \square[/tex]


Sagot :

Let's solve for the missing root step-by-step given the function [tex]\( f(x) = x^2 - 2x - 3 \)[/tex] and the known root [tex]\( x = -1 \)[/tex].

This is a quadratic equation of the form [tex]\( ax^2 + bx + c \)[/tex] where:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = -2 \][/tex]
[tex]\[ c = -3 \][/tex]

To find the roots of the quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

First, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-2)^2 - 4(1)(-3) \][/tex]
[tex]\[ \Delta = 4 + 12 \][/tex]
[tex]\[ \Delta = 16 \][/tex]

Now, the quadratic formula becomes:
[tex]\[ x = \frac{-(-2) \pm \sqrt{16}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{2 \pm 4}{2} \][/tex]

This gives us two solutions (roots):
[tex]\[ x_1 = \frac{2 + 4}{2} = \frac{6}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{2 - 4}{2} = \frac{-2}{2} = -1 \][/tex]

We are given one of the roots [tex]\( x = -1 \)[/tex], so the other root must be:
[tex]\[ x = 3 \][/tex]

Therefore, the missing root is:
[tex]\[ x = 3 \][/tex]
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