Finding Conditions for Real Solutions in Quadratic Equations

The given equation is a quadratic equation: \(x^2 - 4x + 4b = 0\). A quadratic equation is generally represented in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients.

For a quadratic equation to have real solutions, the discriminant must be greater than or equal to zero. The discriminant is denoted by the symbol \(\Delta\) (Delta) and is calculated using the formula: \(\Delta = B^2 - 4AC\), where A, B, and C are the coefficients of the quadratic equation \(Ax^2 + Bx + C = 0\).

Identifying Coefficients

In the given equation \(x^2 - 4x + 4b = 0\), we can identify the coefficients by comparing it to the standard form \(Ax^2 + Bx + C = 0\):

• A = 1 (coefficient of \(x^2\))
• B = -4 (coefficient of x)
• C = 4b (the constant term)

Calculating the Discriminant

Now, we substitute these coefficients into the discriminant formula \(\Delta = B^2 - 4AC\):

\(\Delta = (-4)^2 - 4(1)(4b)\)

\(\Delta = 16 - 16b\)

Applying the Condition for Two Real Solutions

A quadratic equation has two real solutions if the discriminant is greater than or equal to zero (\(\Delta \ge 0\)). So, we set up the inequality:

\(16 - 16b \ge 0\)

Solving the Inequality for 'b'

We need to solve this inequality to find the possible values of 'b'.

Subtract 16 from both sides of the inequality:

\(-16b \ge -16\)

Now, divide both sides by -16. Remember that when you divide or multiply an inequality by a negative number, you must reverse the direction of the inequality sign:

\(\frac{-16b}{-16} \le \frac{-16}{-16}\)

\(b \le 1\)

Conclusion

For the quadratic equation \(x^2 - 4x + 4b = 0\) to have two real solutions, the value of 'b' must be less than or equal to 1. This condition ensures that the discriminant is non-negative.

Detailed Analysis of Options for the Value of 'b'

Let's examine the given options based on our finding that \(b \le 1\) is required for two real solutions:

• Option 1: b = 0. If b=0, then \(0 \le 1\) which is true. The equation \(x^2 - 4x = 0\) becomes \(x(x-4)=0\), giving solutions x=0 and x=4, which are two real solutions.
• Option 2: b ≤ 1. This is exactly the condition we derived. Any value of b satisfying this inequality will result in a non-negative discriminant, guaranteeing two real solutions (either distinct or equal).
• Option 3: b = +1, -1. If b=1, \(1 \le 1\) is true. The equation becomes \(x^2 - 4x + 4 = 0\), which is \((x-2)^2=0\), giving one real solution (a repeated root, often counted as two equal real solutions). If b=-1, \(-1 \le 1\) is true. The equation becomes \(x^2 - 4x - 4 = 0\), which has two distinct real solutions. This option only provides specific values, not the full range.
• Option 4: b ≥ 1. This contradicts our derived condition \(b \le 1\). For example, if b=2, then \(16 - 16(2) = 16 - 32 = -16 < 0\). A negative discriminant means there are no real solutions (only complex/imaginary solutions).

Based on the analysis, the condition \(b \le 1\) correctly describes the range of values for 'b' that yield two real solutions for the given quadratic equation.

Revision Table: Key Concepts

Additional Information: Nature of Quadratic Roots

The discriminant is a powerful tool because it tells us about the nature of the roots of a quadratic equation without actually solving for them. Here’s a bit more detail:

• If \(\Delta > 0\), the equation has two distinct real roots. The graph of the related quadratic function \(y = Ax^2 + Bx + C\) intersects the x-axis at two different points.
• If \(\Delta = 0\), the equation has exactly one real root (sometimes called a double root or repeated root). The graph touches the x-axis at exactly one point.
• If \(\Delta < 0\), the equation has no real roots. It has two complex (or imaginary) roots. The graph does not intersect the x-axis.

In this problem, "two real solutions" means either two distinct real solutions or two equal real solutions. Therefore, we use the condition \(\Delta \ge 0\).