Problem 62
Question
Find the real roots of the equation. \(x^{2}+8 x+16=0\).
Step-by-Step Solution
Verified Answer
The real root of the equation \(x^{2} + 8x + 16 = 0\) is \(x = -4\).
1Step 1: Identify the coefficients of the given equation
The given quadratic equation is in standard form \(ax^{2}+bx+c=0, where \(a=1\), \(b=8\), and \(c=16\) which are the coefficients.
2Step 2: Use the quadratic formula
The quadratic formula is \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Plugging in the given values we have \(x = \frac{-8 \pm \sqrt{(8^{2} - 4*1*16)}}{2*1}\).
3Step 3: Evaluate the expression under the square root
Evaluate \(b^{2} - 4ac\) where \(b=8\), \(a=1\), and \(c=16\). This gives, \(64 - 64 = 0\).
4Step 4: Substitute the evaluated value under the square root into the quadratic formula and solve for x
Using the calculated value, the formula becomes \(x = \frac{-8 \pm \sqrt{0}}{2}\), which simplifies to \(x = -4\) since \(\sqrt{0}=0\).
Key Concepts
Real RootsQuadratic FormulaStandard Form of a Quadratic Equation
Real Roots
In algebra, **real roots** refer to solutions of an equation that are real numbers. These roots appear in quadratic equations when you create a balance in the equation and solve for unknown variables. Real roots are crucial because they represent where a parabola crosses the x-axis on a coordinate plane.
For the equation \(x^2 + 8x + 16 = 0\), the real root calculated is \(-4\). This happens because
For the equation \(x^2 + 8x + 16 = 0\), the real root calculated is \(-4\). This happens because
- The expression under the square root (the discriminant) in the quadratic formula, \(b^2 - 4ac\), equates to zero.
- If the discriminant is zero, the quadratic equation has exactly one real root, because the plus/minus term in the formula eliminates the need for multiple solutions.
Quadratic Formula
The **quadratic formula** is a must-know tool for solving quadratic equations. It is expressed as:
For the equation \(x^2 + 8x + 16 = 0\), substituting the values where \(a = 1\), \(b = 8\), and \(c = 16\), leads to the solution:
The quadratic formula is especially useful because it works even when factoring is difficult or impossible, providing an exact answer swiftly.
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For the equation \(x^2 + 8x + 16 = 0\), substituting the values where \(a = 1\), \(b = 8\), and \(c = 16\), leads to the solution:
- \(x = \frac{-8 \pm \sqrt{64 - 64}}{2}\)
The quadratic formula is especially useful because it works even when factoring is difficult or impossible, providing an exact answer swiftly.
Standard Form of a Quadratic Equation
The **standard form of a quadratic equation** is written as \(ax^2 + bx + c = 0\). This is the general format where:
In the exercise, the equation \(x^2 + 8x + 16 = 0\) is already in standard form, making it easier to identify coefficients:
- \(a\), \(b\), and \(c\) are coefficients, and \(x\) represents the variable.
- \(a\) is not zero to ensure the equation remains quadratic.
In the exercise, the equation \(x^2 + 8x + 16 = 0\) is already in standard form, making it easier to identify coefficients:
- \(a = 1\)
- \(b = 8\)
- \(c = 16\)
Other exercises in this chapter
Problem 61
The vertices of a triangle are \((1,0) .(3,4) .(-1,6)\). Find the point(s) where the medians of this triangle interscet.
View solution Problem 62
Determine the range of \(y=\frac{2 x}{4-x}\) (a) by writing \(y\) in the form \(a+\frac{b}{4-x}\) (b) by first solving the equation for \(x\)
View solution Problem 62
Determine the period. (The least positive number \(p\) for which \(f(x+p)=f(x)\) for all \(x\).) $$f(x)=\sin \frac{1}{2} x$$.
View solution Problem 63
Express the area of a circle as a function of the circumference.
View solution