Question

How do you solve \[w\left( {4w + 6} \right) + 2w = 2\left( {2{w^2} + 7w - 3} \right)\] ?

Answer 1

Hint: After expanding the brackets and using the transposition method we will have a polynomial of degree 2 and it is called quadratic equation or a linear equation. We can solve the quartic using factorization method or by completing the square method or by using quadratic formula.

Complete step-by-step answer:
Given,
 \[w\left( {4w + 6} \right) + 2w = 2\left( {2{w^2} + 7w - 3} \right)\]
Expanding the brackets on both sides we have,
 \[4{w^2} + 6w + 2w = 4{w^2} + 14w - 6\]
Now shifting all the terms to left hand side of the equation we have,
 \[4{w^2} + 6w + 2w - 4{w^2} - 14w + 6 = 0\]
Adding and subtracting the like terms we will have,
 \[ - 6w + 6 = 0\]
Now we transpose 6 to the right hand side of the equation by subtracting 6 on the right hand side,
 \[ - 6w = - 6\]
Dividing -6 on both sides we have
 \[w = \dfrac{{ - 6}}{{ - 6}}\]
 \[ \Rightarrow w = 1\] . This is the required answer.
So, the correct answer is “w = 1”.

Note: We can check whether the obtained solution is correct are not. To do that we substitute the obtained value in he given problem,
 \[w\left( {4w + 6} \right) + 2w = 2\left( {2{w^2} + 7w - 3} \right)\]
 \[1\left( {4\left( 1 \right) + 6} \right) + 2\left( 1 \right) = 2\left( {2{{\left( 1 \right)}^2} + 7\left( 1 \right) - 3} \right)\]
 \[1\left( {4 + 6} \right) + 2 = 2\left( {2 + 7 - 3} \right)\]
 \[1\left( {10} \right) + 2 = 2\left( 6 \right)\]
 \[ \Rightarrow 12 = 12\] . That is LHS is equal to RHS. Hence the obtained solution is correct.