Solving Quadratic Equation Roots and Geometric Progression Problem

We are given two quadratic equations and a condition involving a geometric progression (GP). We need to use the information provided for each equation and the GP relationship to find the required value.

Analyzing the First Quadratic Equation and its Roots

The first equation is \(2x^2 + 3x - \alpha = 0\). Its roots are given as -2 and \(\beta\). For a quadratic equation \(ax^2 + bx + c = 0\), the sum of roots is \(-b/a\) and the product of roots is \(c/a\).

For \(2x^2 + 3x - \alpha = 0\):

• Sum of roots: \((-2) + \beta = -\frac{3}{2}\)
• Product of roots: \((-2) \times \beta = \frac{-\alpha}{2}\)

From the sum of roots, we can find the value of \(\beta\):

\[ -2 + \beta = -\frac{3}{2} \]

\[ \beta = -\frac{3}{2} + 2 \]

\[ \beta = \frac{-3 + 4}{2} = \frac{1}{2} \]

We are given that \(\beta > 0\), and our value \(\beta = 1/2\) satisfies this condition.

Now, using the product of roots, we can find \(\alpha\):

\[ (-2) \times \beta = -\frac{\alpha}{2} \]

Substitute the value of \(\beta\):

\[ (-2) \times \frac{1}{2} = -\frac{\alpha}{2} \]

\[ -1 = -\frac{\alpha}{2} \]

\[ \alpha = 2 \]

We are given that \(\alpha > 0\), and our value \(\alpha = 2\) satisfies this condition.

Understanding the Geometric Progression (GP) Condition

We are told that \(\beta\), 2, and 2m are in GP. In a geometric progression, the ratio between consecutive terms is constant. If a, b, c are in GP, then \(b/a = c/b\), which means \(b^2 = ac\).

Using the terms \(\beta\), 2, 2m, we have:

\[ 2^2 = \beta \times (2m) \]

We already found \(\beta = \frac{1}{2}\). Substitute this value into the GP equation:

\[ 4 = \frac{1}{2} \times (2m) \]

\[ 4 = m \]

So, the value of m is 4.

Examining the Second Quadratic Equation

The second equation is \(x^2 – 3mx + 2m^2 = 0\). We are told that both roots of this equation are positive. Let's substitute the value \(m=4\) into this equation to see if the condition of positive roots holds.

\[ x^2 - 3(4)x + 2(4)^2 = 0 \]

\[ x^2 - 12x + 2(16) = 0 \]

\[ x^2 - 12x + 32 = 0 \]

For a quadratic equation \(Ax^2 + Bx + C = 0\) to have both roots positive, the following conditions must be met:

• The discriminant \(D = B^2 - 4AC\) must be non-negative (\(D \ge 0\)) for real roots.
• The sum of roots (\(-B/A\)) must be positive.
• The product of roots (\(C/A\)) must be positive.

For \(x^2 - 12x + 32 = 0\):

• \(A=1, B=-12, C=32\).
• Discriminant \(D = (-12)^2 - 4(1)(32) = 144 - 128 = 16\). Since \(16 \ge 0\), real roots exist.
• Sum of roots \(= -(-12)/1 = 12\). Since \(12 > 0\), the sum is positive.
• Product of roots \(= 32/1 = 32\). Since \(32 > 0\), the product is positive.

Since the discriminant is non-negative, and the sum and product of roots are positive, both roots of the equation \(x^2 - 12x + 32 = 0\) are indeed positive. This confirms our value of \(m=4\) is consistent with the problem statement.

Calculating the Final Value \({\rm{\beta }}\sqrt {\rm{m}} \)

We have found \(\beta = \frac{1}{2}\) and \(m = 4\). Now we can calculate \({\rm{\beta }}\sqrt {\rm{m}} \):

\[ {\rm{\beta }}\sqrt {\rm{m}} = \frac{1}{2} \sqrt{4} \]

\[ {\rm{\beta }}\sqrt {\rm{m}} = \frac{1}{2} \times 2 \]

\[ {\rm{\beta }}\sqrt {\rm{m}} = 1 \]

The value of \({\rm{\beta }}\sqrt {\rm{m}} \) is 1.

Revision Table: Key Concepts Used

Additional Information: Deep Dive into Concepts

Quadratic Equation Roots: A quadratic equation \(ax^2 + bx + c = 0\) can have real or complex roots depending on the discriminant \(D = b^2 - 4ac\). If \(D \ge 0\), the roots are real. If \(D < 0\), the roots are complex. The sum and product of roots are powerful tools that allow us to analyze root properties without explicitly finding the roots. If the roots are \(r_1\) and \(r_2\), then \(r_1 + r_2 = -b/a\) and \(r_1 r_2 = c/a\).

Geometric Progression (GP): A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For a GP: a, ar, ar², ar³, ... Here, the common ratio is r. If three terms a, b, c are in GP, then b is the geometric mean of a and c, meaning \(b = \sqrt{ac}\) or \(b^2 = ac\). This property is frequently used in problems involving consecutive terms in a GP.

Conditions for Nature of Roots: Consider the quadratic equation \(Ax^2 + Bx + C = 0\):

• Real and Distinct Roots: \(D > 0\)
• Real and Equal Roots: \(D = 0\)
• Complex Conjugate Roots: \(D < 0\)
• Both Roots Positive: \(D \ge 0\), \(-B/A > 0\), \(C/A > 0\)
• Both Roots Negative: \(D \ge 0\), \(-B/A < 0\), \(C/A > 0\)
• Roots have Opposite Signs: \(C/A < 0\) (If \(C/A < 0\), then \(D\) is always positive, as \(D = B^2 - 4AC = B^2 + 4A(-C)\). Since \(A\) and \(C\) have opposite signs, \(AC\) is negative, and \(-4AC\) is positive. So \(B^2 - 4AC\) will be positive plus a non-negative number).

These concepts are fundamental in algebra and frequently appear in competitive exams.