Question

Solve the equation: $2{(x - 3)^2} + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1$

Answer 1

Hint: Here we are asked to solve the given algebraic equation. As we can see in the given equation, we have expressions on both sides of the equation so first we will try to bring all the expressions to one side. Then by simplifying the expression we may get a quadratic equation as we can see that the highest degree is two. Then we will solve that quadratic equation (if any) or by simplifying the expression itself will yield the required answer.

Complete step by step solution:
Given the equation $2{(x - 3)^2} + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1$ and then we need to find the value of the simplified form. We need to find the value of the given variable.
Also, we know that ${(a - b)^2} = (a - b)(a - b) \Rightarrow {a^2} + {b^2} - 2ab$
Thus, using this formula and also using the multiplication operation we get $2{(x - 3)^2} + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1 \Rightarrow 2({x^2} - 6x + 9) + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1$
Again, using the multiplication, we get, $2{x^2} - 12x + 18 + 6{x^2} - 21x + 18 = 8{x^2} - 128 - 1$ since $(x - 2)(2x - 3) = 2{x^2} - 3x - 4x + 6 = 2{x^2} - 7x + 6$
Now using the addition and subtraction operation, we get $8{x^2} - 33x + 36 = 8{x^2} - 129$
By transposition method, let us bring the term \[8{x^2}\] from the left-hand side to the right-hand side and simplify it.
\[ - 33x + 36 = 8{x^2} - 129 - 8{x^2}\]
On simplifying it, we get $ - 33x + 36 = - 129$
Now let us bring the term $36$ to the other side.
$ - 33x = - 129 - 36$
On simplifying the above, we get
$33x = 165$
Hence by the division, we get $x = \dfrac{{165}}{{33}} = 5$
Hence the unknown variable value is $x = 5$ .

Note: In the above problem, we have used a transposition method to simplify. The transposition method is nothing but retaining the unknown variable on one side of the equation and transferring the other terms to the other side of the equation while transferring the terms their signs will change and the operation multiplication will change to division. Also, if we get a quadratic equation $a{x^2} + bx + c = 0$ , we can solve them by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to find its roots.