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Chapter 2: Problem 51

Find the value(s) of k such that the equation has exactly one real root. $$x^{2}+12 x+k=0$$

Short Answer

Expert verified
The value of \(k\) is 36.

Step by step solution

01

Recognize the quadratic equation

The given equation is a quadratic equation in the standard form, which is \(x^2 + 12x + k = 0\). Here, \(a = 1\), \(b = 12\), and \(c = k\).
02

Identify condition for exactly one real root

A quadratic equation has exactly one real root when the discriminant \(b^2 - 4ac\) is equal to zero.
03

Apply the condition for the discriminant

Using the values \(a = 1\), \(b = 12\), and \(c = k\), the discriminant \(b^2 - 4ac = 0\) translates to \(12^2 - 4(1)(k) = 0\).
04

Solve for k

Calculate \(b^2 = 144\), then substitute into the equation: \(144 - 4k = 0\). Solve for \(k\) by adding \(4k\) to both sides and then dividing by 4, getting \(k = 36\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discriminant in Quadratic Equations
The discriminant is a key component in determining the nature and number of roots for a quadratic equation. In the general form of a quadratic equation, \(ax^2 + bx + c = 0\), the discriminant is given by the formula \(b^2 - 4ac\). It is a simple expression that involves the coefficients of the equation: \(a\), \(b\), and \(c\).

  • If the discriminant is positive, the quadratic equation has two distinct real roots.
  • If the discriminant is zero, there is exactly one real root. This situation is also known as a double root or a repeated root.
  • If the discriminant is negative, the quadratic equation has no real roots, but rather two complex conjugate roots.
Understanding the discriminant helps us predict the nature of the solution without actually solving the equation. It's a powerful tool for quickly assessing quadratic equations.
Real Roots
Real roots are solutions to a quadratic equation that can be found on the real number line. The number and type of real roots a quadratic equation has are determined by the value of its discriminant \(b^2 - 4ac\).

  • "One Real Root" occurs when the discriminant is zero, meaning the curve of the quadratic touches the x-axis at exactly one point. This is called a tangent point.
  • "Two Real Roots" happen when the discriminant is greater than zero. Here, the curve crosses the x-axis at two points.
  • "No Real Root" corresponds to a negative discriminant. The curve then remains entirely above or below the x-axis, without touching it.
In the context of our original exercise, we were looking for the value of \(k\) that results in one real root. This occurs when the discriminant equals zero, leading us to solve for \(k\) when the equation was set to \(12^2 - 4k = 0\).
Quadratic Equation Standard Form
Quadratic equations are typically expressed in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a eq 0\). This format is important as it provides a consistent structure for analyzing and solving the equation.

  • The coefficient \(a\) is crucial because if \(a = 0\), the equation becomes linear, not quadratic.
  • \(b\) represents the linear coefficient, influencing the slope of the parabola represented by the equation.
  • \(c\) is the constant term, moving the parabola up or down on a graph.
The standard form not only helps in identifying the coefficients readily but also allows the use of various algebraic methods such as factoring, completing the square, or applying the quadratic formula.
Understanding the standard form is also fundamental in applying the formula for the discriminant, \(b^2 - 4ac\), to understand the nature of the roots of the equation.

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Most popular questions from this chapter

Use the discriminant to determine how many real roots each equation has. $$4 x^{2}-28 x+49=0$$

Determine all of the real-number solutions for each equation. (Remember to check for extraneous solutions.) $$\sqrt{x-5}-\sqrt{x+4}+1=0$$

Show that the quadratic equation $$(x-p)(x-q)=r^{2} \quad(p \neq q)$$ has two distinct real roots.

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$\frac{1}{x-2}-\frac{1}{x-1} \geq \frac{1}{6}$$

Find all real solutions of each equation. For Exercises \(31-36,\) give two forms for each answer: an exact answer (involving a radical) and a calculator approximation rounded to two decimal places. $$(y-1)^{3}=7$$
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