If p, q, r are +ve and are in A.P., the roots of quadratic equation px² + qx + r = 0 are all real fo

Q: Wait, why can we just divide by $r^2$ in the inequality? Doesn't that change the sign?

That is a fantastic question! We can safely divide by $r^2$ because the problem explicitly states that $p, q, r$ are positive. Since $r > 0$, it follows that $r^2$ is strictly positive. Dividing an inequality by a positive number never flips the inequality sign, so our logic remains perfectly sound.

Q: Why do we use the absolute value notation $|t - 7| \ge 4\sqrt{3}$ at the end?

Great observation. The inequality $t^2 - 14t + 1 \ge 0$ gives us two intervals: $t \le 7 - 4\sqrt{3}$ and $t \ge 7 + 4\sqrt{3}$. The expression $|t - 7| \ge 4\sqrt{3}$ is just a mathematically elegant way to combine these two disjoint intervals into a single, compact statement. It tells us that the distance between $t$ and $7$ on the number line must be at least $4\sqrt{3}$.

Q: Could I have solved this by substituting $p$ and $r$ in terms of $q$ instead?

You absolutely could! If you set $p = q - d$ and $r = q + d$, where $d$ is the common difference of the A.P., you would eventually arrive at the same result. However, the method of substituting $q = \frac{p+r}{2}$ is generally faster and keeps the variables $p$ and $r$ directly in the final expression, which matches the options provided.

Question

If $p, q, r$ are +ve and are in A.P., the roots of quadratic equation $px^2 + qx + r = 0$ are all real for

Options:

(A) $|r/p - 7| \ge 4\sqrt{3}$

(B) $|p/r - 7| \ge 4\sqrt{3}$

(C) all $p$ and $r$

(D) no $p$ and $r$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Consider the quadratic equation $px^2 + qx + r = 0$, where $p, q, r$ are positive real numbers. We are given that $p, q, r$ form an Arithmetic Progression (A.P.).

Our objective is to determine the condition on $p$ and $r$ such that the quadratic equation possesses real roots.

### The Discriminant: The Gatekeeper

The nature of the roots of a quadratic equation is determined by its discriminant, $D$. For the roots to be real, we must satisfy the condition $D \ge 0$.

For the given equation, the discriminant is defined as:
$$D = q^2 - 4pr \ge 0$$

If $D < 0$, the parabola lies entirely above the $x$-axis, resulting in complex roots. We must ensure $D \ge 0$ to keep the roots within the real domain.

### The Bridge: The A.P. Constraint

Since $p, q, r$ are in A.P., the middle term $q$ is the arithmetic mean of $p$ and $r$. This provides the relationship:
$$2q = p + r \implies q = \frac{p+r}{2}$$

This substitution allows us to express the discriminant solely in terms of $p$ and $r$. This transformation is the critical step in reducing the complexity of the problem.

### The Algebraic Transformation

Substituting $q = \frac{p+r}{2}$ into the discriminant inequality, we obtain:
$$\left( \frac{p+r}{2} ight)^2 - 4pr \ge 0$$

Expanding the squared term, we get:
$$\frac{p^2 + 2pr + r^2}{4} - 4pr \ge 0$$

Multiplying the entire inequality by $4$ to eliminate the denominator yields:
$$p^2 + 2pr + r^2 - 16pr \ge 0$$
$$p^2 - 14pr + r^2 \ge 0$$

### Final Calculation

To simplify further, we divide the inequality by $r^2$ (since $r > 0$):
$$\left( \frac{p}{r} ight)^2 - 14\left( \frac{p}{r} ight) + 1 \ge 0$$

Let $t = \frac{p}{r}$. We solve the quadratic inequality $t^2 - 14t + 1 \ge 0$ by finding the roots of $t^2 - 14t + 1 = 0$:
$$t = \frac{14 \pm \sqrt{196 - 4}}{2} = 7 \pm \sqrt{48} = 7 \pm 4\sqrt{3}$$

Since the parabola $f(t) = t^2 - 14t + 1$ opens upward, the inequality holds for values outside the roots. Thus, the condition for real roots is:
$$t \le 7 - 4\sqrt{3} \quad 	ext{or} \quad t \ge 7 + 4\sqrt{3}$$

Substituting $t = \frac{p}{r}$ back into the expression, the final condition is:
$$\left| \frac{p}{r} - 7 ight| \ge 4\sqrt{3}$$

**The condition for the existence of real roots is $\left| \frac{p}{r} - 7 ight| \ge 4\sqrt{3}$.**

Animated Solution for Mathematics - Quadratic Equations: If $p, q, r$ are +ve and are in A.P., the roots of quadratic equation $p x^{2} + q x + r = 0$ are all real for

Select Answer:

Visualized Solution

Understanding the Problem

Condition for Real Roots

Using the A.P. Condition

Expanding the Inequality

Simplifying the Expression

Transforming to a Single Variable

Finding the Critical Points

Solving the Inequality

Final Conclusion

The Sigma Insight: Nature of Roots

Analyzing the Setup

The Discriminant

The Bridge

The Algebraic Transformation

Final Calculation

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p,q,r are +ve and are in A.P., the roots of quadratic equation px2+qx+r=0 are all real for