If α and β are the roots of the equation 2x(2x + 1) = 1, then β is equal to : (A) 2α² (B) -2α(α + 1)

Q: Wait, why do we reject the quadratic formula approach?

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ would give you two messy irrational roots. While you *could* technically substitute them into the options, it is computationally expensive and prone to algebraic errors. In JEE Advanced, we look for elegant relationships between roots, not just their numerical values.

Q: Why can't we just use the product of roots?

You absolutely can! The product of roots $\alpha\beta = -\frac{1}{4}$ is a valid relationship. You could write $\beta = -\frac{1}{4\alpha}$. This is another valid path to the answer. However, the sum of roots approach is often more direct when the options are linear or quadratic in $\alpha$.

Q: Why do we substitute $1 = 4\alpha^2 + 2\alpha$?

The goal is to express $\beta$ in terms of $\alpha$ as requested by the options. By substituting the constant $1$ with an expression involving $\alpha$, we effectively 'upgrade' the degree of the expression, allowing us to factorize and simplify it into the form required by the multiple-choice options.

Question

If $\alpha$ and $\beta$ are the roots of the equation $2x(2x + 1) = 1$, then $\beta$ is equal to :

Options:

(A) $2\alpha^2$

(B) $-2\alpha(\alpha + 1)$

(C) $2\alpha(\alpha - 1)$

(D) $2\alpha(\alpha + 1)$

Sigma Percentile · Accepted Answer

### The Elegance of Quadratic Relationships

Imagine standing at the precipice of a complex problem. You see the equation $2x(2x + 1) = 1$ and your first instinct might be to solve for $x$.

But wait! In the world of JEE Advanced, the most direct path is rarely the brute-force path. Let us embark on a journey to uncover the hidden symmetry between the roots $\alpha$ and $\beta$.

### Phase 1: The Standard Form

First, we must bring our equation into the light. Expanding $2x(2x + 1) = 1$ gives us $4x^2 + 2x = 1$.

By subtracting $1$ from both sides, we arrive at the standard quadratic form:

$$4x^2 + 2x - 1 = 0$$

This is the bedrock of our problem. It tells us everything we need to know about the behavior of its roots, $\alpha$ and $\beta$.

### Phase 2: The Bridge Between Roots

We know that for any quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is given by $-\frac{b}{a}$. Applying this to our equation where $a=4$ and $b=2$, we find:

$$\alpha + \beta = -\frac{2}{4} = -\frac{1}{2}$$

This is our bridge. It connects $\alpha$ and $\beta$ in a simple, linear relationship.

If we want to know $\beta$ in terms of $\alpha$, we simply rearrange:

$$\beta = -\frac{1}{2} - \alpha$$

This is a great start, but it doesn't quite match our options, which are quadratic in $\alpha$. We need to go deeper.

### Phase 3: The Power of the Root

Here is the secret weapon of the JEE topper: if $\alpha$ is a root, it *must* satisfy the equation. So, we know that:

$$4\alpha^2 + 2\alpha - 1 = 0$$

If we isolate the constant $1$, we get a powerful substitution:

$$1 = 4\alpha^2 + 2\alpha$$

Now, look back at our expression for $\beta$. We have $\beta = \frac{-1 - 2\alpha}{2}$. If we replace that $1$ with our new expression, we can transform $\beta$ entirely into a function of $\alpha$.

### Phase 4: The Final Synthesis

Let us perform the substitution:

$$\beta = \frac{-(4\alpha^2 + 2\alpha) - 2\alpha}{2}$$

Expanding the numerator, we get $-4\alpha^2 - 2\alpha - 2\alpha = -4\alpha^2 - 4\alpha$. Dividing by $2$, we arrive at:

$$\beta = -2\alpha^2 - 2\alpha$$

Finally, factoring out $-2\alpha$, we reach the elegant conclusion:

$$\beta = -2\alpha(\alpha + 1)$$

This matches our target perfectly. We didn't just solve for a value; we uncovered the structural relationship between the roots.

This is the beauty of algebra—finding the hidden path through the complexity. Keep practicing this, and you will start seeing these patterns everywhere.

Animated Solution for Mathematics - Quadratic Equations: If $α$ and $β$ are the roots of the equation $2 x (2 x + 1) = 1$ , then $β$ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Equation

Expand to Standard Form $a x^{2} + b x + c = 0$

Apply Sum of Roots Property

Isolate $β$ in terms of $α$

Use the Property of a Root

Substitute the Value of $1$

Simplify the Numerator

Divide and Factorize

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Elegance of Quadratic Relationships

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0$ , then $(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Let $- π /6 < θ < - π /12$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

If $p$ and $q$ are the roots of the equation $x^{2} + p x + q = 0$ , then

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $x^{2} + b x + c = 0$ , where $c < 0 < b$ , then

If $α$ and $β$ are the roots of the equation, $7 x^{2} - 3 x - 2 = 0$ , then the value of $\frac{α}{1 - α ^{2}} + \frac{β}{1 - β ^{2}}$ is equal to :

If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 (α^{3} + β^{3})^{2}$ is equal to:

If $α$ and $β$ be two roots of the equation $x^{2} - 64 x + 256 = 0$ . Then the value of $(\frac{α ^{3}}{β ^{5}})^{1/8} + (\frac{β ^{3}}{α ^{5}})^{1/8}$ is

Let $α, β$ be the roots of the equation $x^{2} + 22 x - 1 = 0$ . The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ , is :

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b}$ , is :

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 α and β are the roots of the equation 2x(2x+1)=1, then β is equal to :