Question

Solve the following equation: \[28{x^3} - 9{x^2} + 1 = 0\]

Answer 1

Hint:
Here, we will rewrite the middle term and factorize the equation to get one linear equation and one quadratic equation. We will first solve the linear equation to find one of the roots of the given equation. Then we will solve the quadratic equation using the quadratic formula to get the required roots of the solution.

Formula Used:
Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:
We are given an equation \[28{x^3} - 9{x^2} + 1 = 0\]
Now, we will rewrite the middle term of the equation.
\[ \Rightarrow 28{x^3} - 16{x^2} + 7{x^2} + 1 = 0\]
Now, we will add and subtract \[4x\] on the left hand side, we get
\[ \Rightarrow 28{x^3} - 16{x^2} + 7{x^2} + 4x - 4x + 1 = 0\]
Now, by rearranging the equation, we get
\[ \Rightarrow 28{x^3} - 16{x^2} + 4x + 7{x^2} - 4x + 1 = 0\]
Now, by factoring the equation, we get
\[ \Rightarrow 4x\left( {7{x^2} - 4x + 1} \right) + 1\left( {7{x^2} - 4x + 1} \right) = 0\]
Again factoring out common terms, we get
\[ \Rightarrow \left( {4x + 1} \right)\left( {7{x^2} - 4x + 1} \right) = 0\]
Now applying zero product property, we get
\[ \Rightarrow \left( {4x + 1} \right) = 0\] or \[\left( {7{x^2} - 4x + 1} \right) = 0\]
Now, we consider \[\left( {4x + 1} \right) = 0\]
Subtracting 1 from both sides, we get
\[ \Rightarrow 4x = - 1\]
Dividing both sides by 4, we get
\[ \Rightarrow x = \dfrac{{ - 1}}{4}\]
Now, we will consider \[\left( {7{x^2} - 4x + 1} \right) = 0\]
Comparing the above equation with the general quadratic equation \[a{x^2} + bx + c = 0\] , we get
By substituting \[a = 7,b = - 4\] and \[c = 1\] in the formula \[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get
\[x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 7 \right)\left( 1 \right)} }}{{2\left( 7 \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 28} }}{{14}}\]
By subtracting the terms, we get
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt { - 12} }}{{14}}\]
We know that the square root of a negative number will result in a complex number i.e., \[{i^2} = - 1\]
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {4 \times 3 \times - 1} }}{{14}}\]
\[ \Rightarrow x = \dfrac{{4 \pm 2i\sqrt 3 }}{{14}}\]
By taking out the common factors, we get
\[ \Rightarrow x = \dfrac{{2\left( {2 \pm i\sqrt 3 } \right)}}{{14}}\]
By cancelling out the terms, we get
\[ \Rightarrow x = \dfrac{{\left( {2 \pm i\sqrt 3 } \right)}}{7}\]
By separating the terms, we get
\[ \Rightarrow x = \dfrac{{2 + i\sqrt 3 }}{7};x = \dfrac{{2 - i\sqrt 3 }}{7}\]
\[ \Rightarrow x = \dfrac{2}{7} + i\dfrac{{\sqrt 3 }}{7};x = \dfrac{2}{7} - i\dfrac{{\sqrt 3 }}{7}\]

Therefore, the values of \[x\] are \[\dfrac{{ - 1}}{4},\dfrac{2}{7} + i\dfrac{{\sqrt 3 }}{7}\] and \[\dfrac{2}{7} - i\dfrac{{\sqrt 3 }}{7}\]

Note:
We know that a cubic equation has a highest degree of 3 and has 3 solutions. A quadratic solution has a highest degree of variable as 2 and has only 2 solutions.We know that we can solve the quadratic equation by using any of the four methods. Some quadratic equations cannot be solved by using the factorization method and square root method. But whatever be the quadratic equation, it is quite easy to solve by using the method of quadratic formula. We should be careful that the quadratic equation should be arranged in the right form.