Question

Solve the given quadratic equation: \[4{x^2} - 2\left( {{a^2} + {b^2}} \right)x + {a^2}{b^2} = 0\]

Answer 1

Hint: Here, we have to solve the given quadratic equation. We will factorize the given equation using quadratic formula. Then we will solve the equation further to find the value of the variable. Quadratic equations are polynomial equations with the highest component of degree two.
Formula used: We will use the following formulas:
1.Factorization using Quadratic formula is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , where \[a,b,c\]are the coefficients of \[{x^2},x\] and the constant term respectively.
2.The square of the sum of two numbers is given by the algebraic identity\[{(a + b)^2} = {a^2} + {b^2} + 2ab\].
3.The square of the difference of two numbers is given by the algebraic identity\[{(a - b)^2} = {a^2} + {b^2} - 2ab\].

Complete step-by-step answer:
We are given with an quadratic equation \[4{x^2} - 2\left( {{a^2} + {b^2}} \right)x + {a^2}{b^2} = 0\]
The standard form of a quadratic equation is \[a{x^2} + bx + c = 0\].
By comparing the coefficients of the given quadratic equation with the standard form of quadratic equation, we get
\[\begin{array}{l}a = 4\\b = - 2({a^2} + {b^2})\\c = {a^2}{b^2}\end{array}\]
Now, substituting the values of \[a = 4,b = - 2({a^2} + {b^2})\] and \[c = {a^2}{b^2}\]in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get
\[ \Rightarrow x = \dfrac{{ - \left( { - 2({a^2} + {b^2})} \right) \pm \sqrt {{{\left( { - 2({a^2} + {b^2})} \right)}^2} - 4(4)({a^2}{b^2})} }}{{2(4)}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm \sqrt {4{{({a^2} + {b^2})}^2} - 16({a^2}{b^2})} }}{8}\]
Simplifying the above equation by using the algebraic identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm \sqrt {4({a^4} + {b^4} + 2{a^2}{b^2}) - 16({a^2}{b^2})} }}{8}\]
Multiplying the terms, we get
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm \sqrt {4{a^4} + 4{b^4} + 8{a^2}{b^2} - 16{a^2}{b^2}} }}{8}\]
Subtracting the like terms, we get
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm \sqrt {4{a^4} + 4{b^4} - 8{a^2}{b^2}} }}{8}\]
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm \sqrt {4({a^4} + {b^4} - 2{a^2}{b^2})} }}{8}\]
Now by using algebraic identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow x = \dfrac{{2({a^2} + {b^2}) \pm 2\sqrt {{{({a^2} - {b^2})}^2}} }}{8}\]
\[ \Rightarrow x = \dfrac{{2\left[ {({a^2} + {b^2}) \pm \sqrt {{{({a^2} - {b^2})}^2}} } \right]}}{8}\]
Dividing the numerator and denominator by 2, we get
\[ \Rightarrow x = \dfrac{{\left[ {({a^2} + {b^2}) \pm ({a^2} - {b^2})} \right]}}{4}\]
Writing the terms with positive and negative signs accordingly, we have
\[ \Rightarrow x = \dfrac{{\left[ {({a^2} + {b^2}) + ({a^2} - {b^2})} \right]}}{4}\] or \[x = \dfrac{{\left[ {({a^2} + {b^2}) - ({a^2} - {b^2})} \right]}}{4}\]
Adding and subtracting the like terms, we have
\[ \Rightarrow x = \dfrac{{2{a^2}}}{4}\] or \[x = \dfrac{{2{b^2}}}{4}\]
Dividing the numerator and denominator by 2, we get
\[ \Rightarrow x = \dfrac{{{a^2}}}{2}\] or \[x = \dfrac{{{b^2}}}{2}\]
Therefore, we get the values as \[x = \dfrac{{{a^2}}}{2}\] and \[x = \dfrac{{{b^2}}}{2}\] .

Note: Here, we are provided with the quadratic equation. Quadratic equation is an equation which has a highest degree of 2. We should be careful that the quadratic equation is in the right form while using the quadratic formula. When the equation can’t be reduced using the factorization method and completing the square method, this method is used. This method is applicable to all quadratic equations.