The quadritic equations x² - 6x + a = 0 and x² - cx + 6 = 0 have one root in common. The other roots

Q: Wait, why can't we just use the discriminant to solve this?

The discriminant is a powerful tool for determining the nature of roots (real, imaginary, equal), but it doesn't give us the specific values of the roots. Here, we need the actual values to satisfy the ratio and integer constraints. Vieta's formulas are much more direct for relating roots to coefficients.

Q: Why did we define the other roots as $4k$ and $3k$?

The problem states the other roots are in the ratio $4:3$. Any two numbers in a ratio $a:b$ can be written as $ak$ and $bk$ for some constant $k$. This is a standard algebraic technique to reduce the number of unknowns.

Q: Why do we reject $\alpha = 4$ even though it's a valid root of the quadratic?

Mathematics is about satisfying all constraints. While $\alpha = 4$ is a solution to the quadratic equation $\alpha^2 - 6\alpha + 8 = 0$, it fails the 'integer' constraint for the other roots. If $\alpha = 4$, the other root of the second equation becomes $1.5$, which is not an integer. Therefore, it must be discarded.

Question

The quadritic equations $x^2 - 6x + a = 0$ and $x^2 - cx + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

Options:

(A) 1

(B) 4

(C) 3

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let us define our variables. We have the common root $\alpha$.

For the first equation, $x^2 - 6x + a = 0$, the roots are $\alpha$ and $4k$.

For the second equation, $x^2 - cx + 6 = 0$, the roots are $\alpha$ and $3k$. The ratio of the other roots is $4:3$, which serves as our bridge between the two equations.

The critical constraint is that $4k$ and $3k$ must be integers. This is the filter through which our final answer must pass.

### The Power of Vieta's Formulas

Now, we invoke the wisdom of Vieta. For the first equation, the sum of the roots is given by:

$$\alpha + 4k = 6$$

This is our first anchor.

Next, we look at the second equation, $x^2 - cx + 6 = 0$. We know the product of the roots is the constant term $6$:

$$\alpha \cdot 3k = 6$$

Simplifying this, we obtain:

$$\alpha k = 2$$

This is our second anchor.

### The Algebraic Dance

We now have a system of equations: $\alpha + 4k = 6$ and $\alpha k = 2$.

From the first equation, we can write $4k = 6 - \alpha$, or $k = \frac{6 - \alpha}{4}$.

Substituting this into our second equation, we get:

$$\alpha \left( \frac{6 - \alpha}{4} \right) = 2$$

Multiplying by $4$, we obtain $\alpha(6 - \alpha) = 8$, which expands to $6\alpha - \alpha^2 = 8$.

Rearranging this, we arrive at the quadratic:

$$\alpha^2 - 6\alpha + 8 = 0$$

### The Final Verdict

Factoring this quadratic is straightforward:

$$(\alpha - 2)(\alpha - 4) = 0$$

This gives us two candidates: $\alpha = 2$ or $\alpha = 4$.

We must now return to our integer constraint.

If $\alpha = 2$, then $k = \frac{6 - 2}{4} = 1$. The other roots are $4(1) = 4$ and $3(1) = 3$. Since both are integers, this is a valid solution.

If $\alpha = 4$, then $k = \frac{6 - 4}{4} = 0.5$. The other roots are $4(0.5) = 2$ and $3(0.5) = 1.5$.

Since $1.5$ is not an integer, this solution is invalid. Thus, the only possible common root is **$\alpha = 2$**.

Animated Solution for Mathematics - Quadratic Equations: The quadritic equations $x^{2} - 6 x + a = 0$ and $x^{2} - c x + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

Select Answer:

Visualized Solution

Defining the Roots

Applying Sum of Roots to Equation 1

Applying Product of Roots to Equation 2

Expressing $k$ in terms of $α$

Substituting $k$ into the Product Equation

Simplifying to a Quadratic Equation

Solving for $α$

Verifying the Integer Constraint

Final Conclusion

The Sigma Insight: Common Roots

Analyzing the Setup

The Power of Vieta's Formulas

The Algebraic Dance

The Final Verdict

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STATEMENT-1: $(p^{2} - q) (b^{2} - a c) \geq 0$ and STATEMENT-2: $b \neq = p a$ or $c \neq = q a$

Select Answer:

Visualized Solution

Defining the Roots

Applying Sum of Roots to Equation 1

Applying Product of Roots to Equation 2

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k into the Product Equation

Simplifying to a Quadratic Equation

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

Verifying the Integer Constraint

Final Conclusion

The Sigma Insight: Common Roots

Analyzing the Setup

The Power of Vieta's Formulas

The Algebraic Dance

The Final Verdict

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