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?

• A. \( |h_{1}|<h_{2} \)
• B. \( |h_{1}|=|h_{2}| \)
• C. \( |h_{1}|>h_{2} \)
• D. \( \frac{h_{1}}{h_{2}}>-1 \)

Hint:

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!

Answer 1

The Correct Option is C

Step 1: Understand diminishing roots.
To lower each root by $h$, we replace $x$ with $y+h$. The new polynomial in $y$ has its $y^2$ coefficient equal to $\dfrac{f''(h)}{2}$.
Step 2: Set the condition.
We want no $x^2$ term, so that coefficient must be zero. This means $f''(h) = 0$.
Step 3: Find the second derivative.
From $f(x)=x^4+3x^3-7x^2+4x+1$, we get $f'(x)=4x^3+9x^2-14x+4$, then $f''(x)=12x^2+18x-14$.
Step 4: Solve for h.
Set $12h^2+18h-14=0$ and divide by 2 to get $6h^2+9h-7=0$. The quadratic formula gives \[ h = \frac{-9 \pm \sqrt{249}}{12} \]
Step 5: Name the two roots by sign.
The negative one is $h_1 = \dfrac{-9-\sqrt{249}}{12}$ and the positive one is $h_2 = \dfrac{-9+\sqrt{249}}{12}$.
Step 6: Compare their sizes.
Now $|h_1| = \dfrac{9+\sqrt{249}}{12}$ and $h_2 = \dfrac{\sqrt{249}-9}{12}$. Since $9+\sqrt{249} > \sqrt{249}-9$, we conclude \[ \boxed{|h_1| > h_2} \]