Question

How do you find the real solutions of polynomials $14{x^2} + 5 = 3{x^4}$?

Answer 1

Hint:
In this question, we need to find the real solutions of the given polynomial. Firstly, we will bring all the terms in the R.H.S. to the L.H.S. to solve the polynomial. We substitute $t = {x^2}$. Then we obtain a quadratic polynomial in terms of the variable t. Then we solve this quadratic equation using quadratic formula and obtain the roots. Then we substitute back $t = {x^2}$ and then solve to obtain the required solution.

Complete step by step solution:
Given the polynomial of the form $14{x^2} + 5 = 3{x^4}$ …… (1)
We are asked to find the real solutions for the above polynomial given by the equation (1).
Firstly, bring all terms in the R.H.S. of the equation to L.H.S. and proceed further.
So transferring $3{x^4}$ to the L.H.S. we get,
$ \Rightarrow - 3{x^4} + 14{x^2} + 5 = 0$
Multiplying by -1 throughout the equation, we get,
$ \Rightarrow - 1( - 3{x^4} + 14{x^2} + 5) = - 1(0)$
Simplifying we get,
$ \Rightarrow 3{x^4} - 14{x^2} - 5 = 0$
This can also be written as,
$ \Rightarrow 3{({x^2})^2} - 14{x^2} - 5 = 0$ …… (2)
Now we take $t = {x^2}$ and simplify further to obtain the solution.
Hence, the equation (2) becomes,
$ \Rightarrow 3{t^2} - 14t - 5 = 0$
Note that this is a quadratic equation. We now solve it to obtain the roots.
If $a{x^2} + bx + c = 0$ is a quadratic equation, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Note that here $a = 3$, $b = - 14$ and $c = - 5$.
Hence we get,
$t = \dfrac{{ - ( - 14) \pm \sqrt {{{( - 14)}^2} - 4 \times 3 \times ( - 5)} }}{{2 \times 3}}$
$ \Rightarrow t = \dfrac{{14 \pm \sqrt {196 + 60} }}{6}$
$ \Rightarrow t = \dfrac{{14 \pm \sqrt {256} }}{6}$
We know the value of $\sqrt {256} = 16$
Hence we get,
$ \Rightarrow t = \dfrac{{14 \pm 16}}{6}$
Now substituting back $t = {x^2}$ we get,
$ \Rightarrow {x^2} = \dfrac{{14 \pm 16}}{6}$
When ${x^2} = \dfrac{{14 + 16}}{6}$
$ \Rightarrow {x^2} = \dfrac{{30}}{6}$
$ \Rightarrow {x^2} = 5$
Taking square root on both sides we get,
$ \Rightarrow \sqrt {{x^2}} = \sqrt 5 $
$ \Rightarrow x = \pm \sqrt 5 $, which is the real solution.
When ${x^2} = \dfrac{{14 - 16}}{6}$
$ \Rightarrow {x^2} = \dfrac{{ - 2}}{6}$
$ \Rightarrow {x^2} = \dfrac{{ - 1}}{3}$
Taking square root on both sides, we get,
$ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{{ - 1}}{3}} $
We know that $i = \sqrt { - 1} $, which is an imaginary unit.
Hence we have,
$ \Rightarrow \sqrt {{x^2}} = \sqrt { - 1} \times \sqrt {\dfrac{1}{3}} $
$ \Rightarrow x = \pm i\sqrt {\dfrac{1}{3}} $, which is not a real solution.
i.e. it is an imaginary or complex solution.

Hence, the real solution of the polynomial $14{x^2} + 5 = 3{x^4}$ is given by $x = \pm \sqrt 5 $.

Note:
Students may get confused with the term real solution of the polynomial. Remember that the real solution is the one which is free from the imaginary unit $i = \sqrt { - 1} $. If any solution contains the imaginary unit, then it is an imaginary or complex solution.
Also remember the quadratic formula to find out the roots of the quadratic polynomial.
If $a{x^2} + bx + c = 0$ is a quadratic equation, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.