Question

Find the roots of the following equation $35{{y}^{2}}+\dfrac{12}{{{y}^{2}}}=44$ then:
A. \[y=\pm \sqrt{\dfrac{6}{7}},\pm \sqrt{\dfrac{2}{5}}\]
B. \[y=\pm \sqrt{\dfrac{6}{11}},\pm \sqrt{\dfrac{2}{5}}\]
C. \[y=\pm \sqrt{\dfrac{6}{7}},\pm \sqrt{\dfrac{2}{7}}\]
D. \[y=\pm \sqrt{\dfrac{6}{7}},\pm \sqrt{\dfrac{1}{5}}\]

Answer 1

Hint: We can solve this question by converting it into a quadratic equation. If the converted equation can be factored then we can solve it by splitting middle term otherwise we can solve it by quadratic formula.

Complete step-by-step solution -
Given equation is $35{{y}^{2}}+\dfrac{12}{{{y}^{2}}}=44$
We can write equation as
$\dfrac{35{{y}^{4}}+12}{{{y}^{2}}}=44$
$35{{y}^{4}}+12=44{{y}^{2}}$
$35{{y}^{4}}-44{{y}^{2}}+12=0$
We can let ${{y}^{2}}=t$, then we can write equation as:
 $35{{t}^{2}}-44t+12=0$
Now it is a quadratic equation. We can factorize it by splitting the middle term. In this method we split middle terms in such two parts whose summation is equal to middle term and product equal to product of first and last term. In the above equation the first term is $35{{t}^{2}}$ and last term is 12.
Product of first and last term is $35{{t}^{2}}\times 12=420{{t}^{2}}$
Middle term of the equation is $-44t$.
So we need to break $-44t$in such two parts whose sum is equal to $-44t$ and product equal to $420{{t}^{2}}$.
Hence we can write it as $-44t=-30t-14t$because multiplication of $30t$ and$14t$ equal to $420{{t}^{2}}$.
Hence equation is
$\Rightarrow 35{{t}^{2}}-30t-14t+12=0$
$\Rightarrow 5t(7t-6)-2(7t-6)=0$
$\Rightarrow (7t-6)(5t-2)=0$
Either $7t-6=0$
$t=\dfrac{6}{7}$
Or $5t-2=0$
$t=\dfrac{2}{5}$
As we let ${{y}^{2}}=t$
So we can write either ${{y}^{2}}=\dfrac{6}{7}$ or ${{y}^{2}}=\dfrac{2}{5}$
We can solve it as $y=\pm \sqrt{\dfrac{6}{7}}$or $y=\pm \sqrt{\dfrac{2}{5}}$
Hence possible values of y are $\dfrac{6}{7},\dfrac{-6}{7},\dfrac{2}{5},\dfrac{-2}{5}$.

Note: As we can see the degree (highest exponent of variable in equation) of the converted equation is 4 but we can convert it into quadratic. This type of equation is known as the Biquadratic equation. In general, the number of solutions of any equation is equal to the degree of that equation. That’s why we get 4 possible values of y. So in this question we don’t need to forget to put back the value of t and then solve for y.