Let a, b, c be real. If ax² + bx + c =...

MathematicsLocation of RootsJEE Advanced 1995Easy

Let $a, b, c$ be real. If $a x^{2} + b x + c = 0$ has two real roots $α$ and $β$ , where $α < - 1$ and $β > 1$ , then show that $1 + c / a + ∣ b / a ∣ < 0$ .

Visualized Solution (Hindi)

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x)

Let the quadratic equation be $a x^{2} + b x + c = 0$ .
Divide by $a$ to define $f (x) = x^{2} + \frac{b}{a} x + \frac{c}{a}$ .
Since the coefficient of $x^{2}$ is $1 > 0$ , the parabola opens upwards.

Visualizing the Root Constraints

Given roots $α$ and $β$ such that $α < - 1$ and $β > 1$ .
This implies the interval $[- 1, 1]$ lies entirely between the roots $α$ and $β$ .
For an upward-opening parabola, $f (x) < 0$ for all $x \in (α, β)$ .

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (1) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (- 1)

Since $- 1, 1 \in (α, β)$ , we must have $f (1) < 0$ and $f (- 1) < 0$ .
Substituting $x = 1$ : $1 + \frac{b}{a} + \frac{c}{a} < 0$ .
Substituting $x = - 1$ : $1 - \frac{b}{a} + \frac{c}{a} < 0$ .

Combining with Absolute Value

We have: $1 + \frac{c}{a} + \frac{b}{a} < 0$ and $1 + \frac{c}{a} - \frac{b}{a} < 0$ .
Recall that $∣ x ∣ = max (x, - x)$ .
Therefore, $1 + \frac{c}{a} + \frac{b}{a} < 0$ must hold true.

Conclusion and Key Takeaway

Key Takeaway: If an interval $[k_{1}, k_{2}]$ lies between the roots of $f (x) = 0$ , then $a \cdot f (k) < 0$ for any $k \in [k_{1}, k_{2}]$ .
Final Result: $1 + \frac{c}{a} + \frac{b}{a} < 0$ is proved.
Next Challenge: What if only one root was in the interval $(- 1, 1)$ ? How would the condition change?

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Conceptually Similar Problems

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

(A)

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(B)

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(C)

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(D)

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Answer:B

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(A)

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(B)

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(C)

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(D)

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Answer:D

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If $x^{2} + (a - b) x + (1 - a - b) = 0$ where $a, b \in R$ then find the values of $a$ for which equation has unequal real roots for all values of $b$ .

Answer:

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Let $a > 0, b > 0$ and $c > 0$ . Then the roots of the equation $a x^{2} + b x + c = 0$

(A)

are real and negative

(B)

have negative real parts

(C)

both (a) and (b)

(D)

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Let $a, b, c$ be the sides of a triangle where $a \neq = b \neq = c$ and $λ \in R$ . If the roots of the equation $x^{2} + 2 (a + b + c) x + 3 λ (ab + b c + c a) = 0$ are real, then

(A)

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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 $α, β$ .

Answer:

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MathematicsMaximum and Minimum Values of Quadratic ExpressionsJEE Main 2009Easy

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If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

(A)

less than

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(B)

greater than

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(C)

less than

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(D)

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If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

MathematicsGeometrical Applications of Complex NumbersJEE Main 2011Easy

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Let $α, β$ be real and $z$ be a complex number. If $z^{2} + α z + β = 0$ has two distinct roots on the line $R ez = 1$ , then it is necessary that

(A)

β \in (- 1, 0)

(B)

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(C)

β \in (1, \infty)

(D)

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Answer:C

Let $a, b, c$ be real. If $a x^{2} + b x + c = 0$ has two real roots $α$ and $β$ , where $α < - 1$ and $β > 1$ , then show that $1 + c / a + ∣ b / a ∣ < 0$ .

Visualized Solution (Hindi)

Defining the Function $f (x)$

Visualizing the Root Constraints

Evaluating $f (1)$ and $f (- 1)$

Combining with Absolute Value

Conclusion and Key Takeaway

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be real. If ax2+bx+c=0 has two real roots α and β, where α<−1 and β>1, then show that 1+c/a+∣b/a∣<0.

Visualized Solution (Hindi)

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Visualizing the Root Constraints

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−1)

Combining with Absolute Value

Conclusion and Key Takeaway

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Let $a > 0, b > 0$ and $c > 0$ . Then the roots of the equation $a x^{2} + b x + c = 0$

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