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Magazine Chapter 3: Problem 37
Use the quadratic formula to find exact solutions. $$x^{2}-2 x=15$$
Short Answer
Expert verified
x = 5 or x = -3
Step by step solution
01
Write the equation in standard form
Rearrange the given equation to match the standard form of a quadratic equation, which is \(ax^2 + bx + c = 0\). For the given equation:\(x^2 - 2x = 15\), subtract 15 from both sides to get:\(x^2 - 2x - 15 = 0\)
02
Identify coefficients
Identify the values of \(a\), \(b\), and \(c\) from the standard form of the quadratic equation \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -2\), and \(c = -15\)
03
Write down the quadratic formula
The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
04
Substitute coefficients into the formula
Substitute \(a = 1\), \(b = -2\), and \(c = -15\) into the quadratic formula:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}\]
05
Simplify under the square root
Evaluate the expression under the square root (the discriminant): \[(-2)^2 - 4(1)(-15) = 4 + 60 = 64\]
06
Calculate the roots
With the simplified discriminant, now we have:\[x = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}\]This results in two solutions:For the positive case: \[x = \frac{2 + 8}{2} = \frac{10}{2} = 5\]For the negative case: \[x = \frac{2 - 8}{2} = \frac{-6}{2} = -3\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic equation
A quadratic equation is a polynomial equation of the second degree. This means that the highest power of the variable (usually represented as x) is 2. The general form of a quadratic equation is:
\[ ax^2 + bx + c = 0 \]
Here:
\[ ax^2 + bx + c = 0 \]
Here:
- a is the coefficient of \(x^2\)
- b is the coefficient of x
- c is the constant term
discriminant
The discriminant is a key part of the quadratic formula, represented by the expression under the square root:
\[ b^2 - 4ac \]
The value of the discriminant gives important information about the nature of the roots of a quadratic equation:
\( (-2)^2 - 4(1)(-15) = 4 + 60 = 64 \)
Since the discriminant is positive (64), we can conclude that the quadratic equation \(x^2 - 2x - 15 = 0\) has two distinct real roots.
\[ b^2 - 4ac \]
The value of the discriminant gives important information about the nature of the roots of a quadratic equation:
- If the discriminant is positive (\resizebox{6pt}{6pt}{+}), the quadratic equation has two distinct real roots.
- If it is zero (\resizebox{6pt}{6pt}{0}), the quadratic equation has exactly one real root (this root is repeated).
- If the discriminant is negative (\resizebox{6pt}{6pt}{-}), the quadratic equation has two complex roots (not real roots).
\( (-2)^2 - 4(1)(-15) = 4 + 60 = 64 \)
Since the discriminant is positive (64), we can conclude that the quadratic equation \(x^2 - 2x - 15 = 0\) has two distinct real roots.
roots of a quadratic equation
The roots of a quadratic equation are the values of x that satisfy the equation. Using the quadratic formula, we can find these roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our specific problem, we substitute \(a = 1\), \(b = -2\), and \(c = -15\) into the formula:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{64}}{2} \]
This simplifies to:
\[ x = \frac{2 \pm 8}{2} \]
This results in two solutions:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our specific problem, we substitute \(a = 1\), \(b = -2\), and \(c = -15\) into the formula:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{64}}{2} \]
This simplifies to:
\[ x = \frac{2 \pm 8}{2} \]
This results in two solutions:
- For the positive case: \( x = \frac{2 + 8}{2} = 5 \)
- For the negative case: \( x = \frac{2 - 8}{2} = -3 \)
