Question

Solve the following quadratic equation using quadratic formula: $a\left(x^2+1\right)=x\left(a^2+1\right)$
a). $\\{ \dfrac{1}{a},a\\}$
b). $\\{ \dfrac{1}{a},1\\}$
c). $\\{ \dfrac{1}{a},{a^2}\\}$
d). None of these

Answer 1

Use the technique of taking terms common from the given equation and then solve, then write the equation into simpler expressions and forms. Then after writing it into forms of factors in multiplied form, equate it to the right hand side of the equation as given and get the solution.

Complete step-by-step solution:
We are given with the quadratic equation
Let us simplify the equation into simpler form and then use the quadratic formula to get the value of $x$
Let us reduce the equation into simplest form possible i.e.
Now we use the quadratic formula or the Sridharacharya method to find the value of $x$i.e.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here $a = a,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} b = - ({a^2} + 1),{\kern 1pt} {\kern 1pt} c = a$then,
$x = \dfrac{{({a^2} + 1) \pm \sqrt {{{({a^2} + 1)}^2} - 4{a^2}} }}{{2a}}$
Now let us simplify the terms inside the square root to simplest form i.e.

{({a^2} + 1)^2} - 4{a^2} \\\ = {({a^2})^2} + {1^2} + 2.{a^2}.1 - 4{a^2} \\\ = {a^4} + 1 + 2{a^2} - 4{a^2} \\\ = {a^4} - 2{a^2} + 1 \\\ = {({a^2})^2} - 2.{a^2}.1 + {1^2} \\\ = {({a^2} - 1)^2} $$ Now we substitute the value inside the square root to get our desired result,

x = \dfrac{{({a^2} + 1) \pm \sqrt {{{({a^2} - 1)}^2}} }}{{2a}} \\
\Rightarrow x = \dfrac{{({a^2} + 1) \pm ({a^2} - 1)}}{{2a}} \\
\Rightarrow x = \dfrac{{({a^2} + 1) + ({a^2} - 1)}}{{2a}},x = \dfrac{{({a^2} + 1) - ({a^2} - 1)}}{{2a}} $$\Rightarrow x = \dfrac{{2{a^2}}}{{2a}},x = \dfrac{2}{{2a}}$$\Rightarrow x = a,x = \dfrac{1}{a}$**Therefore, our correct option is (a)$\{ \dfrac{1}{a},a\} $$.**
Additional information: In mathematics, a quadratic is a sort of example that offers a variable extended via itself-an operation called squaring. This language derives from the vicinity of a square being its side length increased by way of itself. The word "quadratic" comes from quadratum, the Latin word for square. Quadratic equations are virtually used in everyday life, as whilst calculating regions, figuring out a product's earnings or formulating the velocity of an item. Quadratic equations consult with equations with at least one squared variable, with the most standard form being. Quadratic equations are used to represent different functions, different real life processes and used in various fields like chemistry, biology, physics and many other fields to solve and get the desired solution.

Note: It is important that we know various ways, techniques, shortcuts and methods to solve different types of quadratic equations. Middle term factoring, Sridharacharya method or Quadratic formula, graphical methods are some techniques of solving the roots of a quadratic equation.