The value of 'a' for which one root of the quadratic equation (a² - 5a + 3)x² + (3a - 1)x + 2 = 0 is

Q: Wait, why can't we just use the quadratic formula to find the roots and then set one equal to twice the other?

That is a valid instinct, but it is a trap! Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ would introduce a massive square root term. Setting $\frac{-B + \sqrt{D}}{2A} = 2 \left( \frac{-B - \sqrt{D}}{2A} \right)$ would require you to square both sides, likely leading to algebraic errors and a much more complex equation. Vieta's formulas allow us to bypass the roots themselves and work directly with the coefficients, which is the elegant, professional way to solve this.

Q: What if the denominator $a^2 - 5a + 3$ is zero? Do we need to check that?

Excellent observation! You are thinking like a mathematician. If $a^2 - 5a + 3 = 0$, the equation is no longer quadratic; it becomes a linear equation, and the concept of 'two roots' changes. In this problem, we assume the equation is quadratic, so we implicitly assume $a^2 - 5a + 3 \neq 0$. When we cancel this term from both sides, we are effectively assuming it is non-zero. If you were to find a value of $a$ that makes the denominator zero, you would have to reject it.

Q: Why did we square the sum relation? Is there another way?

Squaring the sum relation is the most direct way to create an $\alpha^2$ term, which allows us to substitute it directly into the product relation. You could also solve for $\alpha$ in the sum relation and substitute it into the product relation, but squaring is generally cleaner. The goal is to eliminate the variable $\alpha$ because it is an auxiliary variable we introduced. Any method that eliminates $\alpha$ and leaves you with an equation in $a$ is perfectly valid!

Question

The value of 'a' for which one root of the quadratic equation $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$ is twice as large as the other is

Options:

(A) $-1/3$

(B) 2/3

(C) $-2/3$

(D) 1/3

Sigma Percentile · Accepted Answer

### The Symphony of Roots: A Journey into Quadratics

Welcome, future engineer. Today, we are not just solving a quadratic equation; we are uncovering the hidden symmetry within a mathematical structure.

When you look at the equation $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$, it is easy to feel overwhelmed by the parameter $a$. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Wisdom of Vieta

We are told that one root is twice the other. Let us name our roots $\alpha$ and $2\alpha$. This is our starting point.

We do not need to know the exact values of $x$ to find $a$. Instead, we turn to the masters of algebra: Vieta's formulas. These formulas are the bridge between the roots of a polynomial and its coefficients.

For any quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $-\frac{B}{A}$ and the product is $\frac{C}{A}$.

In our specific case, our coefficients are $A = a^2 - 5a + 3$, $B = 3a - 1$, and $C = 2$. By applying Vieta's, we immediately gain two powerful insights:

1. The Sum: 
$$\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3}$$

2. The Product: 
$$\alpha \cdot 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3}$$

Look at these two equations. They are the keys to the kingdom. We have two unknowns, $\alpha$ and $a$, and two equations. This is a solvable system.

### Phase 2: The Algebraic Bridge

Now, we must eliminate $\alpha$. This is where the strategy comes in. From the sum relation, we can isolate $\alpha$:

$$3\alpha = \frac{1 - 3a}{a^2 - 5a + 3}$$

Dividing by 3 gives us $\alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)}$. This looks a bit messy, but do not fear.

We have a product relation that involves $\alpha^2$. If we square our expression for $\alpha$, we will have a direct match for the product relation. Let us square both sides of our sum relation:

$$\alpha^2 = \left[ \frac{1 - 3a}{3(a^2 - 5a + 3)} \right]^2 = \frac{(1 - 3a)^2}{9(a^2 - 5a + 3)^2}$$

This is the moment of truth. We now have two different expressions for $\alpha^2$. One from the product relation, and one from the squared sum relation. By setting them equal, we create a single equation that depends only on $a$.

### Phase 3: The Grand Cancellation

Equating our two expressions for $\alpha^2$ gives us:

$$\frac{1}{a^2 - 5a + 3} = \frac{(1 - 3a)^2}{9(a^2 - 5a + 3)^2}$$

Take a deep breath. Look at the denominators. We have $(a^2 - 5a + 3)$ on the left and $(a^2 - 5a + 3)^2$ on the right.

We can multiply both sides by $(a^2 - 5a + 3)$ to simplify this significantly. This is the beauty of algebra—the complexity collapses when you see the structure. We are left with:

$$1 = \frac{(1 - 3a)^2}{9(a^2 - 5a + 3)}$$

Cross-multiplying brings us to a much friendlier linear-quadratic hybrid:

$$9(a^2 - 5a + 3) = (1 - 3a)^2$$

### Phase 4: The Final Resolution

Now, we expand both sides. On the left, we distribute the 9: $9a^2 - 45a + 27$. On the right, we expand the square of the binomial: $1 - 6a + 9a^2$.

Watch closely—the $9a^2$ terms on both sides are identical. They cancel out! This is not a coincidence; it is the mathematical universe rewarding your persistence.

We are left with a simple linear equation:

$$-45a + 27 = 1 - 6a$$

Rearranging the terms to solve for $a$:

$$27 - 1 = 45a - 6a$$
$$26 = 39a$$
$$a = \frac{26}{39}$$

Simplifying this fraction by dividing both numerator and denominator by 13, we arrive at our final answer: **$a = \frac{2}{3}$**.

### Conclusion

See? What started as a daunting quadratic equation with a parameter $a$ was dismantled piece by piece using the elegant tools of Vieta's formulas and careful algebraic manipulation. You didn't just solve a problem; you navigated a logical path.

Keep this mindset—look for the structure, trust the process, and the answer will always reveal itself. You are doing great.

Animated Solution for Mathematics - Quadratic Equations: The value of 'a' for which one root of the quadratic equation $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice as large as the other is

Select Answer:

Visualized Solution

Visualizing the Roots: $α$ and $2 α$

Sum of Roots: $3 α = - \frac{B}{A}$

Product of Roots: $2 α^{2} = \frac{C}{A}$

Expressing $α$ from the Sum Relation

Squaring $α$ to Match the Product Relation

Equating the Two Expressions for $α^{2}$

Simplifying by Canceling Common Terms

Expanding and Solving for $a$

Final Calculation: $a = \frac{2}{3}$

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Roots

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

If the difference between the roots of the equation $x^{2} + a x + 1 = 0$ is less than $5$ , then the set of possible values of $a$ is

The value of $a$ for which the sum of the squares of the roots of the equation $x^{2} - (a - 2) x - a - 1 = 0$ assume the least value is

If $α, β$ are the roots of the equation $x^{2} - (5 + 3^{l o g_{3} 5} - 5^{l o g_{5} 3}) x + 3 (3^{(l o g_{3} 5)^{1/3}} - 5^{(l o g_{5} 3)^{2/3}} - 1) = 0$ then the equation, whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ , is

The numbers of pairs $(a, b)$ of real numbers, such that whenever $α$ is a root of the equation $x^{2} + a x + b = 0$ , $α^{2} - 2$ is also a root of this equation, is :

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to

Let $f (x)$ be a quadratic polynomial such that $f (- 1) + f (2) = 0$ . If one of the roots of $f (x) = 0$ is 3, then its other root lies in :

Let $a, b \in R, a \neq = 0$ , such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation $x^{2} - 2 b x - 10 = 0$ . If $β$ is the root of this equation, then $α^{2} + β^{2}$ is equal to:

If $λ$ be the ratio of the roots of the quadratic equation in $x$ , $3 m^{2} x^{2} + m (m - 4) x + 2 = 0$ , then the least value of $m$ for which $λ + 1/ λ = 1$ , is :

Let $α, β$ be the roots of the equation $x^{2} - 2 x + 6 = 0$ and $\frac{1}{α ^{2}} + 1, \frac{1}{β ^{2}} + 1$ be the roots of the equation $x^{2} + a x + b = 0$ . Then the roots of the equation $x^{2} - (a + b - 2) x + (a + b + 2) = 0$ are :

Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

Select Answer:

Visualized Solution

Visualizing the Roots: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2α

Sum of Roots: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3α=−AB​

Product of Roots: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2α2=AC​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α from the Sum Relation

Squaring .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α to Match the Product Relation

Equating the Two Expressions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2