Question

To remove the second term of the equation ${x^4} - 10{x^3} + 35{x^2} - 50x + 24 = 0$, diminish the roots by
A) $\dfrac{2}{5}$
B) $- \dfrac{2}{5}$
C) $\dfrac{5}{2}$
D) $- \dfrac{5}{2}$

Answer 1

Firstly, put $x = y + h$, in the given equation.
Then, expand each term and get an equation in the form of $a{y^4} + b{y^3} + c{y^2} + dy + e = 0$.
Now, to remove the second term, the coefficient of second term, i.e. $b = 0$.
Thus, get the required answer.

Complete step by step solution:
The equation given here is ${x^4} - 10{x^3} + 35{x^2} - 50x + 24 = 0$ .
Now, replacing $x = y + h$ , we get
${\left( {y + h} \right)^4} - 10{\left( {y + h} \right)^3} + 35{\left( {y + h} \right)^2} - 50\left( {y + h} \right) + 24 = 0$
Now, solving the above equation as follows
${\left( {{y^2} + {h^2} + 2yh} \right)^2} - 10\left( {{y^3} + {h^3} + 3y{h^2} + 3{y^2}h} \right) + 35\left( {{y^2} + {h^2} + 2yh} \right) - 50y - 50h + 24 = 0$
$\therefore {\left( {{y^2}} \right)^2} + {\left( {{h^2}} \right)^2} + {\left( {2yh} \right)^2} + 2\left( {{y^2}} \right)\left( {{h^2}} \right) + 2\left( {{h^2}} \right)\left( {2yh} \right) + 2\left( {2yh} \right)\left( {{y^2}} \right) - 10{y^3} - 10{h^3} - 30y{h^2} - \\\ 30{y^2}h + 35{y^2} + 35{h^2} + 70yh - 50y - 50h + 24 = 0 \\\$
$\therefore {y^4} + {h^4} + 4{y^2}{h^2} + 2{y^2}{h^2} + 4y{h^3} + 4{y^3}h - 10{y^3} - 10{h^3} - 30y{h^2} - 30{y^2}h + 35{y^2} + 35{h^2} + \\\ 70yh - 50y - 50h + 24 = 0 \\\$
Now, writing the terms in the form of $a{y^4} + b{y^3} + c{y^2} + dy + e = 0$
$\therefore {y^4} + {y^3}\left( {4h - 10} \right) + {y^2}\left( {4{h^2} + 2{h^2} - 30h + 35} \right) + y\left( {4{h^3} - 30{h^2} + 70h - 50} \right) + \left( {{h^4} - 10{h^3} + 35{h^2} - 50h + 24} \right) = 0$ As it is said that, the second term of the equation, must be removed.
So, to remove the second term of the equation, the coefficient of the second term must be 0.
$\therefore 4h - 10 = 0 \\\ \therefore 4h = 10 \\\ \therefore h = \dfrac{{10}}{4} \\\ \therefore h = \dfrac{5}{2} \\\$
Thus, we get the value of h as $h = \dfrac{5}{2}$ .
Now, the reduced equation can be written as

$$\therefore {y^4} + {y^3}\left( {4\left( {\dfrac{5}{2}} \right) - 10} \right) + {y^2}\left( {4{{\left( {\dfrac{5}{2}} \right)}^2} + 2{{\left( {\dfrac{5}{2}} \right)}^2} - 30\left( {\dfrac{5}{2}} \right) + 35} \right) + y\left( {4{{\left( {\dfrac{5}{2}} \right)}^3} - 30{{\left( {\dfrac{5}{2}} \right)}^2} + 70\left( {\dfrac{5}{2}} \right) - 50} \right) + \\\ \left( {{{\left( {\dfrac{5}{2}} \right)}^4} - 10{{\left( {\dfrac{5}{2}} \right)}^3} + 35{{\left( {\dfrac{5}{2}} \right)}^2} - 50\left( {\dfrac{5}{2}} \right) + 24} \right) = 0 \\\$$

Thus, the reduced equation is ${y^4} - \dfrac{5}{2}{y^2} - 211 = 0$.
We can also write the above equation as ${x^4} - \dfrac{5}{2}{x^2} - 211 = 0$
Thus, to remove the second term, we have to diminish the roots by $\dfrac{5}{2}$ .

So, option (C) is correct.

Note:
Here, the expansion of the terms must be calculated carefully using appropriate formulae.
The formulae used in the expansion in the question are:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\\ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b \\\ {\left( {a + b} \right)^4} = {\left[ {{{\left( {a + b} \right)}^2}} \right]^2} = {\left( {{a^2} + {b^2} + 2ab} \right)^2} \\\$