Question

Solve the following Quadratic Equation
$\bold{\texttt{Using Formula(Sridhara Acharyya's method)}}\\\tt x=\dfrac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}$
$\large \bold{Question : - } \\ \orange{ \tt \: a( {x}^{2} + 1) = ( {a}^{2} + 1)x \green{ \: \: \: , a \neq \: 0}}$
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Answer 1

Topic :-

Quadratic Equation

Given :-

$\mathtt{a(x^2+1)=(a^2+1)x}$

$\mathtt{a \neq 0}$

To Find :-

Value of 'x' using Quadratic formula which is commonly known as Shreedhara Acharya's formula.

Formula to be Used :-

Shreedhara Acharya's formula

Value of 'x' for a quadratic equation Ax² + Bx + C = 0 is given by,

$\mathtt{x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}}$

Solution :-

$\mathtt{a(x^2+1)=(a^2+1)x}$

Open Brackets,

$\mathtt{ax^2+a=a^2x+x}$

Transpose (a²x + x) in LHS,

$\mathtt{ax^2+a-a^2x-x=0}$

Rearrange terms,

$\mathtt{ax^2-a^2x-x+a=0}$

$\mathtt{ax^2-(a^2+1)x+a=0}$

Compare this equation with general form of Quadratic Equation : Ax² + Bx + C = 0,

$\mathtt{We\:can\:observe\:that,}$

$\mathtt{A=a}$

$\mathtt{B=-(a^2+1)}$

$\mathtt{C=a}$

Apply formula,

$\mathtt{x=\dfrac{-(-(a^2+1))\pm\sqrt{(-(a^2+1))^2-4(a)(a)}}{2a}}$

$\mathtt{x=\dfrac{(a^2+1)\pm\sqrt{(a^2+1)^2-4a^2}}{2a}}$

$\mathtt{(\because (-(a^2+1))^2=(a^2+1)^2)}$

$\mathtt{x=\dfrac{(a^2+1)\pm\sqrt{a^4+2a^2+1-4a^2}}{2a}}$

$\mathtt{(\because (x+y)^2=x^2+2xy+y^2)}$

$\mathtt{x=\dfrac{(a^2+1)\pm\sqrt{a^4-2a^2+1}}{2a}}$

$\mathtt{x=\dfrac{(a^2+1)\pm\sqrt{(a^2-1)^2}}{2a}}$

$\mathtt{(\because (x-y)^2=x^2-2xy+y^2)}$

$\mathtt{x=\dfrac{(a^2+1)\pm(a^2-1)}{2a}}$

Case 1,

Take positive sign,

$\mathtt{x=\dfrac{(a^2+1)+(a^2-1)}{2a}}$

$\mathtt{x=\dfrac{a^2+a^2+1-1}{2a}}$

$\mathtt{x=\dfrac{2a^2}{2a}}$

$\mathtt{x=a}$

Case 2,

Take negative sign,

$\mathtt{x=\dfrac{(a^2+1)-(a^2-1)}{2a}}$

$\mathtt{x=\dfrac{a^2+1-a^2+1}{2a}}$

$\mathtt{x=\dfrac{a^2-a^2+1+1}{2a}}$

$\mathtt{x=\dfrac{2}{2a}}$

$\mathtt{x=\dfrac{1}{a}}$

Answer :-

So, value of x is

$\mathtt{a\;or\;\dfrac{1}{a}.}$

Answer 2

Question:-

Using the formula :-

$\\ \sf{ x=\dfrac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}}$

Solve the following Quadratic Equation:-

$\\ \sf{ \bold{\: a( {x}^{2} + 1) = ( {a}^{2} + 1)x { \: \: \: , a \neq \: 0}}}$

Required Answer:-

Given Equation:-

$\\ \sf{ \rightarrow{\: a( {x}^{2} + 1) = ( {a}^{2} + 1)x }}$

To Find:-

$\\ \sf{ \rightarrow{the \: value \: of \: \bold{x}}}$

♡Solution:-

$\\ \sf{ \bold{\: a( {x}^{2} + 1) = ( {a}^{2} + 1)x }}$

$\\ \sf{ \implies{a {x}^{2} + a = ( {a}^{2} + 1)x}}$

$\\ \sf{ \implies{a {x}^{2} + a - ( {a}^{2} + 1)x = 0}}$

$\\ \sf{ \implies{a {x}^{2} - ( {a}^{2} + 1)x + a = 0}}$

$\\ \sf{ \implies{a {x}^{2} + ( - a {}^{2} - 1)x + a = 0}}$

$\\ \rm{Comparing \: this \: equation \: with \: \bold{a {x}^{2} + bx + c = 0} \: we \: find:-}$

$\\ \sf{ \bold{a} = a \: , \: \: \bold{b} = ( - {a}^{2} - 1) \: and \: \bold{c} = a}$

As we know that;

$\\ \boxed{ \sf{ \blue{ x=\dfrac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}}}}$

Substituting the values of a, b and c :-

$\\ \sf{ \bold{x} = \frac{ - ( - {a} ^{2} - 1) \pm \sqrt{( - {a}^{2} - 1) {}^{2} - 4 \times a \times a } }{2a} }$

Now,

$\\ \rm{Simplifying \: \sqrt{( - {a}^{2} - 1) {}^{2} - 4 \times a \times a }:-}$

$\\ \sf{ \bold{\sqrt{( - {a}^{2} - 1) {}^{2} - 4 \times a \times a }}}$

$\\ \sf{ \implies{\sqrt{( - {a}^{2}) {}^{2} - 2 \times {a}^{2} \times {1}^{2} - 4 {a}^{2} }}}$

$\\ \sf{ \implies{\sqrt{ {a}^{4} - 2 {a}^{2} + 1 }}}$

$\\ \sf{ \implies{\sqrt{ ( {a}^{2} ) {}^{2} - 2 \times {a}^{2} \times 1 + {1}^{2} }}}$

$\\ \sf{ \implies{\sqrt{ ( {a}^{2} - 1) {}^{2} }}}$

$\\ \sf{ \implies{ {(a)}^{2} - (1) {}^{2} }}$

$\\ \sf{ \implies{ \bold{(a + 1)(a - 1)}}}$

$\\ \\ \rm{ \therefore{ \bold {\: \sqrt{( - {a}^{2} - 1) {}^{2} - 4 \times a \times a } = (a + 1)(a - 1)}}}$

Therefore,

$\\ \sf{ \bold{x} = \frac{ - ( - a {}^{2} - 1) \pm(a + 1)(a - 1) }{2a} }$

Hence,

$\\ \sf{ \bold{{x}_{1}} = \frac{ - ( - a {}^{2} - 1) + (a + 1)(a - 1)}{2a} }$

$\\ \sf{ \bold{{x}_{1}} = \frac{ {a}^{2} + 1 + {a}^{2} - 1}{2a} }$

$\\ \sf{ \bold{{x}_{1}} = \frac{ 2 {a}^{2} }{2a} }$

$\\ \sf{ \therefore{ \bold{ \red{{x}_{1} = a}}}}$

Again,

$\\ \sf{ \bold{{x}_{2}} = \frac{ - ( - a {}^{2} - 1) - (a + 1)(a - 1)}{2a} }$

$\\ \sf{ \bold{{x}_{2}} = \frac{ {a}^{2} + 1 - {a}^{2} + 1}{2a} }$

$\\ \sf{ \bold{{x}_{2}} = \frac{ 2 }{2a} }$

$\\ \sf{ \therefore{ \bold{ \red{{x}_{2} = \frac{1}{a} }}}}$

$\\ \\ \underline{ \boxed{ \rm{ \green{Hence,\: value \: of \: \bold{x = a \: , \: \frac{1}{a} }}}}}$