A value of b for which the equations x² + bx - 1 = 0, x² + x + b = 0 have one root in common is (A)

Q: Why can't we just use the discriminant to find the common root?

The discriminant ($D = b^2 - 4ac$) only tells us about the nature of the roots (real, imaginary, equal). It does not directly tell us if two different quadratic equations share a root. For common roots, we must use the algebraic condition of satisfying both equations simultaneously.

Q: Why did we subtract the equations instead of adding them?

Subtracting is a strategic move! Both equations have an $x^2$ term. By subtracting, we eliminate the $x^2$ term entirely, reducing the system to a linear equation in $x$. This makes it much easier to express $x$ (or $\alpha$) in terms of $b$.

Q: Wait, why is the answer an imaginary number? Did I make a mistake?

Not at all! In JEE Advanced problems, coefficients and roots can certainly be complex numbers. The fact that we arrived at $b^2 = -3$ is perfectly valid. Always trust your algebraic steps if your logic is sound.

Question

A value of $b$ for which the equations $x^2 + bx - 1 = 0, x^2 + x + b = 0$ have one root in common is

Options:

(A) $-\sqrt{2}$

(B) $-i\sqrt{3}$

(C) $i\sqrt{5}$

(D) $\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast, abstract landscape where two quadratic equations, $x^2 + bx - 1 = 0$ and $x^2 + x + b = 0$, are waiting for you. They share a secret: a common root, which we shall call $\alpha$.

If $\alpha$ is a root, it must satisfy both equations simultaneously. This gives us our starting point:

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

$$\alpha^2 + \alpha + b = 0$$

### The Power of Elimination

Both equations contain an $\alpha^2$ term. If we subtract the second equation from the first, the $\alpha^2$ terms will vanish, leaving us with a much simpler, linear relationship.

Performing the subtraction:

$$(\alpha^2 + b\alpha - 1) - (\alpha^2 + \alpha + b) = 0$$

The $\alpha^2$ terms cancel out, leaving us with $(b - 1)\alpha - (1 + b) = 0$. With a quick rearrangement, we find the expression for the common root:

$$\alpha = \frac{b + 1}{b - 1}$$

### The Algebraic Grind

Now, we substitute $\alpha = \frac{b + 1}{b - 1}$ into the second equation, $x^2 + x + b = 0$. This yields:

$$\left(\frac{b + 1}{b - 1}\right)^2 + \left(\frac{b + 1}{b - 1}\right) + b = 0$$

To clear the denominators, we multiply the entire equation by $(b - 1)^2$:

$$(b + 1)^2 + (b + 1)(b - 1) + b(b - 1)^2 = 0$$

### The Final Reveal

Expanding each term with care, we obtain:

$$(b^2 + 2b + 1) + (b^2 - 1) + b(b^2 - 2b + 1) = 0$$

$$b^2 + 2b + 1 + b^2 - 1 + b^3 - 2b^2 + b = 0$$

Combining like terms, the $b^2$ terms and the constants cancel out, leaving us with the elegant cubic equation:

$$b^3 + 3b = 0$$

Factoring this expression, we get $b(b^2 + 3) = 0$. This leads to the solutions $b = 0$ or $b^2 = -3$.

Since $b^2 = -3$ implies $b = \pm i\sqrt{3}$, we have identified the possible values for the parameter. The **final solutions** for $b$ are **$0, i\sqrt{3}, -i\sqrt{3}$**.

Animated Solution for Mathematics - Quadratic Equations: A value of $b$ for which the equations $x^{2} + b x - 1 = 0, x^{2} + x + b = 0$ have one root in common is

Select Answer:

Visualized Solution

Define the Common Root $α$

Eliminate $α^{2}$ by Subtraction

Substitute $α$ back into Equation (2)

Clear the Denominator

Expand and Simplify the Terms

Final Polynomial in $b$

Solve for $b$ and Match Options

The Sigma Insight: Common Roots

Analyzing the Setup

The Power of Elimination

The Algebraic Grind

The Final Reveal

Similar Questions

If the quadratic equations $x^{2} + a x + b = 0$ and $x^{2} + b x + a = 0$ ( $a \neq = b$ ) have a common root, then the numerical value of $a + b$ is .........

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real roots is $\frac{3}{2 β}$ , then $β$ is equal to

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

If the equations $x^{2} + 2 x + 3 = 0$ and $a x^{2} + b x + c = 0, a, b, c \in R$ , have a common root, then $a : b : c$ is

If $α, β$ are the roots of $x^{2} + p x + q = 0$ and $γ, δ$ are the roots of $x^{2} + r x + s = 0$ , evaluate $(α - γ) (α - δ) (β - γ) (β - δ)$ in terms of $p, q, r$ and $s$ . Deduce the condition that the equations have a common root.

Let $λ \neq = 0$ be in $R$ . If $α$ and $β$ are the roots of the equation $x^{2} - x + 2 λ = 0$ , and $α$ and $γ$ are the roots of equation $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to .

If $α, β$ and $γ$ are three consecutive terms of a non-constant G.P. such that the equations $x^{2} + 2 α x + β = 0$ and $x^{2} + β x - 1 = 0$ have a common root, then $α (β + γ)$ is equal to :

If $a, b, c$ are in G.P., then the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root if $d / a, e / b, f / c$ are in

Let $λ \neq = 0$ be in $R$ . If $α$ and $β$ are the roots of the equation, $x^{2} - x + 2 λ = 0$ and $α$ and $γ$ are the roots of the equation, $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to :

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

Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A value of b for which the equations x2+bx−1=0,x2+x+b=0 have one root in common is

Select Answer:

Visualized Solution

Define the Common Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Eliminate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2 by Subtraction

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α back into Equation (2)

Clear the Denominator

Expand and Simplify the Terms

Final Polynomial in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

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