Question:
The equation formed with the roots obtained by diminishing the roots of the equation \( x^{4}+3x^{3}-7x^{2}+4x+1=0 \) by 'h', does not contain the \( x^{2} \) term. If the possible values of such 'h' are \( h_{1}<0 \) and \( h_{2}>0, \) then which one of the following is true?
The equation formed with the roots obtained by diminishing the roots of the equation \( x^{4}+3x^{3}-7x^{2}+4x+1=0 \) by 'h', does not contain the \( x^{2} \) term. If the possible values of such 'h' are \( h_{1}<0 \) and \( h_{2}>0, \) then which one of the following is true?
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For any quadratic equation \( Ax^2 + Bx + C = 0 \), if the sum of the roots \( -\frac{B}{A} \) is negative, the negative root will always have a larger absolute magnitude than the positive root. Here, sum = \( -\frac{9}{6} = -1.5 \), so \( |h_1| > h_2 \) immediately!
Updated On: Jun 7, 2026
- \( |h_{1}|<h_{2} \)
- \( |h_{1}|=|h_{2}| \)
- \( |h_{1}|>h_{2} \)
- \( \frac{h_{1}}{h_{2}}>-1 \)
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The Correct Option is C
Solution and Explanation
Concept:
To diminish the roots of a polynomial \( f(x) \) by \( h \), we shift the variable by substituting \( x = y + h \). The new polynomial in terms of \( y \) will have its coefficients determined by Taylor's expansion or synthetic division:
\[
f(y+h) = f(h) + f'(h)y + \frac{f''(h)}{2!}y^2 + \frac{f'''h)}{3!}y^3 + \dots
\]
For the transformed equation to lack the \( x^2 \) (or \( y^2 \)) term, the coefficient of \( y^2 \) must be zero, which implies \( f''(h) = 0 \).
Step 1: Finding the second derivative of the given polynomial.
Let \( f(x) = x^{4}+3x^{3}-7x^{2}+4x+1 \). First derivative: \[ f'(x) = 4x^3 + 9x^2 - 14x + 4 \] Second derivative: \[ f''(x) = 12x^2 + 18x - 14 \]
Step 2: Solving for \( h \) using \( f''(h) = 0 \).
Set \( 12h^2 + 18h - 14 = 0 \). Dividing the entire equation by 2: \[ 6h^2 + 9h - 7 = 0 \] Using the quadratic formula \( h = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \): \[ h = \frac{-9 \pm \sqrt{9^2 - 4(6)(-7)}}{2(6)} = \frac{-9 \pm \sqrt{81 + 168}}{12} = \frac{-9 \pm \sqrt{249}}{12} \] Thus, the two roots are: \[ h_1 = \frac{-9 - \sqrt{249}}{12} < 0, \quad h_2 = \frac{-9 + \sqrt{249}}{12} > 0 \]
Step 3: Comparing the absolute values \( |h_1| \) and \( h_2 \).
Let us find the absolute values: \[ |h_1| = \frac{9 + \sqrt{249}}{12} \] \[ |h_2| = h_2 = \frac{\sqrt{249} - 9}{12} \] Comparing the two numerators, it is absolutely clear that \( 9 + \sqrt{249} > \sqrt{249} - 9 \). Therefore: \[ |h_1| > h_2 \]
Step 1: Finding the second derivative of the given polynomial.
Let \( f(x) = x^{4}+3x^{3}-7x^{2}+4x+1 \). First derivative: \[ f'(x) = 4x^3 + 9x^2 - 14x + 4 \] Second derivative: \[ f''(x) = 12x^2 + 18x - 14 \]
Step 2: Solving for \( h \) using \( f''(h) = 0 \).
Set \( 12h^2 + 18h - 14 = 0 \). Dividing the entire equation by 2: \[ 6h^2 + 9h - 7 = 0 \] Using the quadratic formula \( h = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \): \[ h = \frac{-9 \pm \sqrt{9^2 - 4(6)(-7)}}{2(6)} = \frac{-9 \pm \sqrt{81 + 168}}{12} = \frac{-9 \pm \sqrt{249}}{12} \] Thus, the two roots are: \[ h_1 = \frac{-9 - \sqrt{249}}{12} < 0, \quad h_2 = \frac{-9 + \sqrt{249}}{12} > 0 \]
Step 3: Comparing the absolute values \( |h_1| \) and \( h_2 \).
Let us find the absolute values: \[ |h_1| = \frac{9 + \sqrt{249}}{12} \] \[ |h_2| = h_2 = \frac{\sqrt{249} - 9}{12} \] Comparing the two numerators, it is absolutely clear that \( 9 + \sqrt{249} > \sqrt{249} - 9 \). Therefore: \[ |h_1| > h_2 \]
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