Question

How do you solve the algebraic equation \[3{\left( x \right)^2} - 6x + 7 = {x^2} + 3x - x\left( {x + 1} \right) + 3\]?

Answer 1

Hint: To solve any quadratic equation, the first step is to simplify and write the equation in standard form and then apply algebraic identities if possible. Next step is to choose an appropriate method for obtaining \[x\] either quadratic formula or middle term splitting method. Therefore, simplify the given quadratic equation and then solve the equation by the use of quadratic formula.
The roots for the quadratic equation \[a{x^2} + bx + c = 0\] can be written as \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step solution:
Simplify the given equation as shown below.
\[3{\left( x \right)^2} - 6x + 7 = {x^2} + 3x - x\left( {x + 1} \right) + 3\]
\[ \Rightarrow 3{x^2} - 6x + 7 = {x^2} + 3x - {x^2} - x + 3\]

Eliminate the term \[{x^2}\] from the right hand side as shown below.
\[ \Rightarrow 3{x^2} - 6x + 7 = 3x - x + 3\]

Simplify the right hand side of the equation as shown below.
\[ \Rightarrow 3{x^2} - 6x + 7 = 3x - x + 3\]
\[ \Rightarrow 3{x^2} - 6x + 7 = 2x + 3\]

Subtract 3 from both sides of the equation as shown below.
\[ \Rightarrow 3{x^2} - 6x + 7 - 3 = 2x + 3 - 3\]
\[ \Rightarrow 3{x^2} - 6x + 4 = 2x\]

Subtract \[2x\] from both sides of the equation and simplify as shown below.
\[ \Rightarrow 3{x^2} - 6x + 4 - 2x = 2x - 2x\]
\[ \Rightarrow 3{x^2} - 8x + 4 = 0\]

Use the quadratic formula to obtain the solution of the above equation as follows:

Substitute \[3\] as \[a\], \[ - 8\] as \[b\] and \[4\] as \[c\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] as shown below.
\[x = \dfrac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( 3 \right)\left( 4 \right)} }}{{2\left( 3 \right)}}\]

Simplify each term in the expression as shown below.
\[ \Rightarrow x = \dfrac{{8 \pm \sqrt {64 - 48} }}{6}\]
\[ \Rightarrow x = \dfrac{{8 \pm \sqrt {16} }}{6}\]
\[ \Rightarrow x = \dfrac{{8 \pm 4}}{6}\]

Therefore, two solutions of the equation are real and can be expressed as shown below.
\[ \Rightarrow x = \dfrac{{8 + 4}}{6}\,\,or\,\,\dfrac{{8 - 4}}{6}\]

Simply each solution as follows:
\[ \Rightarrow x = \dfrac{{12}}{6}\,\,or\,\,\dfrac{4}{6}\]
\[ \Rightarrow x = 2\,\,or\,\,\dfrac{2}{3}\]

Thus, two solutions for the given equation are \[2\] and \[\dfrac{2}{3}\].

Note: The alternative method for the solution of the given equation is obtained by the use of the middle term splitting method as shown below.
\[3{\left( x \right)^2} - 6x + 7 = {x^2} + 3x - x\left( {x + 1} \right) + 3\]

Simplify the equation and write the equation in standard form of quadratic equation as shown below.
\[3{x^2} - 6x + 7 = {x^2} + 3x - {x^2} - x + 3\]
\[ \Rightarrow 3{x^2} - 6x + 7 = 3x - x + 3\]
\[ \Rightarrow 3{x^2} - 6x + 7 = 2x + 3\]

Subtract \[2x\] and \[3\] from both sides of the equation as shown below.
\[ \Rightarrow 3{x^2} - 6x + 7 - 2x - 3 = 0\]
\[ \Rightarrow 3{x^2} - 8x + 4 = 0\]

Now use the method of middle term splitting to solve the obtained quadratic equation as follows:
\[ \Rightarrow 3{x^2} - \left( {6 + 2} \right)x + 4 = 0\]
\[ \Rightarrow 3{x^2} - 6x - 2x + 4 = 0\]
\[ \Rightarrow 3x\left( {x - 2} \right) - 2\left( {x - 2} \right) = 0\]
\[ \Rightarrow \left( {x - 2} \right)\left( {3x - 2} \right) = 0\]

Therefore, to satisfy the equation either \[x - 2 = 0\] or \[3x - 2 = 0\].
\[ \Rightarrow x = 2\,\,or\,\,x = \dfrac{2}{3}\]

Thus, two solutions for the given equation are \[2\] and \[\dfrac{2}{3}\].