If 2 and 6 are the roots of the equation ax² + bx + 1 = 0, then the quadratic equation, whose roots

Q: Wait, why can't I just substitute $x=2$ and $x=6$ into the new expression directly?

Because those are the roots of the original equation, not the new one! The new equation is a transformation. You must find the specific values of the new roots first to build the new quadratic.

Q: Is there a faster way than finding $a$ and $b$?

You could try to manipulate the expressions algebraically, but in this case, finding $a$ and $b$ is the most robust path. Don't fear the arithmetic; embrace the clarity it brings.

Q: Why is the final equation $x^2 + 8x + 12 = 0$ and not something else?

Because the sum of the new roots is $-8$ and the product is $12$. The standard form $x^2 - (Sum)x + (Product) = 0$ dictates the signs. Since the sum is negative, the middle term becomes positive.

Question

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

Options:

(A) $2x^2 + 11x + 12 = 0$

(B) $x^2 + 8x + 12 = 0$

(C) $4x^2 + 14x + 12 = 0$

(D) $x^2 + 10x + 16 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a detective, and the quadratic equation $ax^2 + bx + 1 = 0$ is a locked safe. We have two clues: the roots are $2$ and $6$.

In the world of JEE Advanced, we use the elegant tools of Vieta's formulas. These formulas are the bridge between the roots and the coefficients.

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

### Unlocking the Coefficients

Applying this to our equation, the sum of the roots is $2 + 6 = 8$, which must equal $-\frac{b}{a}$. Simultaneously, the product is $2 \cdot 6 = 12$, which must equal $\frac{1}{a}$.

From the product, we immediately see that:

$$12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$$

With $a$ in hand, we return to the sum equation:

$$8 = -\frac{b}{a} \Rightarrow b = -8a$$

Substituting $a = \frac{1}{12}$, we get:

$$b = -8 \cdot \left( \frac{1}{12} ight) = -\frac{8}{12} = -\frac{2}{3}$$

### The Transformation: Calculating the New Roots

Now that we have our coefficients $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, we face the second part of our journey: the transformation. We are asked to find a new quadratic equation with roots $r_1 = \frac{1}{2a+b}$ and $r_2 = \frac{1}{6a+b}$.

Let us calculate $r_1$ first. The denominator is:

$$2a + b = 2\left( \frac{1}{12} ight) - \frac{2}{3} = \frac{1}{6} - \frac{4}{6} = -\frac{3}{6} = -\frac{1}{2}$$

Therefore, $r_1 = \frac{1}{-1/2} = -2$.

Now for $r_2$. The denominator is:

$$6a + b = 6\left( \frac{1}{12} ight) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}$$

Therefore, $r_2 = \frac{1}{-1/6} = -6$.

### The Final Construction: Building the Equation

We are at the finish line. To construct a quadratic equation from its roots, we use the standard template: $x^2 - (	ext{Sum of Roots})x + (	ext{Product of Roots}) = 0$.

The sum of our new roots is $S = (-2) + (-6) = -8$. The product of our new roots is $P = (-2) \cdot (-6) = 12$.

Plugging these into our template, we get:

$$x^2 - (-8)x + 12 = 0$$

Simplifying the signs, we arrive at the final result:

**$x^2 + 8x + 12 = 0$**

Animated Solution for Mathematics - Quadratic Equations: 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 :

Select Answer:

Visualized Solution

Given Equation & Roots

Sum of Roots

Product of Roots

Solving for $a$ and $b$

Defining the New Roots

Calculating First New Root $r_{1}$

Calculating Second New Root $r_{2}$

Sum and Product of New Roots

Forming the Final Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Unlocking the Coefficients

The Transformation

The Final Construction

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Select Answer:

Visualized Solution

Given Equation & Roots

Sum of Roots

Product of Roots

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Defining the New Roots

Calculating First New Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​

Calculating Second New Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​