Question

How do you solve ${(4x - 3)^2} - {(2x - 1)^2} = 0$?

Answer 1

Hint: Use the formula of ${(a - b)^2} = {a^2} + {b^2} - 2ab$ to expand the squares and then simplify according to the Question given. After Expanding, we would get similar terms. So, just add or subtract the following and solve further to get the values of x by equating them with zero.

Complete step-by-step answer:
The given equation is ${(4x - 3)^2} - {(2x - 1)^2} = 0$ …………(i)
Let’s separately solve it:
Take ${(4x - 3)^2}$ and expand it using the formula of ${(a - b)^2} = {a^2} + {b^2} - 2ab$:
$
  {(4x - 3)^2} = ({(4x)^2} + {(3)^2} - 2 \times 4x \times 3) \\
   = (16{x^2} + 9 - 24x) \\
   = (16{x^2} - 24x + 9) \;
 $ ……….(ii)
Take ${(2x - 1)^2}$ and expand it using the formula of ${(a - b)^2} = {a^2} + {b^2} - 2ab$:
$
  {(2x - 1)^2} = ({(2x)^2} + {(1)^2} - 2 \times 2x \times 1) \\
   = (4{x^2} + 1 - 4x) \\
   = (4{x^2} - 4x + 1) \;
 $ ………..(iii)
Put (ii) and (iii) in (i):
  $
  {(4x - 3)^2} - {(2x - 1)^2} = 0 \\
  (16{x^2} - 24x + 9) - (4{x^2} + 1 - 4x) = 0 \\
 $
Solve the above Equation:
$
  {(4x - 3)^2} - {(2x - 1)^2} = 0 \\
  (16{x^2} - 24x + 9) - (4{x^2} + 1 - 4x) = 0 \\
  (16{x^2} - 24x + 9 - 4{x^2} - 1 + 4x) = 0 \\
  (12{x^2} - 20x + 8) = 0 \;
 $
Take $4$ from $(12{x^2} - 20x + 8)$ as it is the highest common factor, which would result:
 $(12{x^2} - 20x + 8) = 4(3{x^2} - 5x + 2)$
On Equating with $0$, we get:
$(3{x^2} - 5x + 2) = 0$
We got a Quadratic Equation, so we will further solve it using the Quadratic Equation Formula that is:
\[
  x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \;
  \] and $x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$x_1,x_2$are root to quadratic equation $a{x^2} + bx + c$
On comparing $a{x^2} + bx + c$ and $(3{x^2} - 5x + 2)$,we get:
$a = 3$,$b = - 5$and $c = 2$
Put the Values in the above formula of $x_1,x_2$:
\[
  x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \\
   = \dfrac{{ - ( - 5) + \sqrt {{{( - 5)}^2} - 4 \times 3 \times 2} }}{{2 \times 3}} \\
   = \dfrac{{5 + \sqrt {25 - 24} }}{6} \\
   = \dfrac{{5 + \sqrt 1 }}{6} \\
   = \dfrac{{5 + 1}}{6} \\
   = \dfrac{6}{6} = 1 \;
 \]
And, for $x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
\[
   = \dfrac{{ - ( - 5) - \sqrt {{{( - 5)}^2} - 4 \times 3 \times 2} }}{{2 \times 3}} \\
   = \dfrac{{5 - \sqrt {25 - 24} }}{6} \\
   = \dfrac{{5 - \sqrt 1 }}{6} \\
   = \dfrac{{5 - 1}}{6} \\
   = \dfrac{4}{6} = \dfrac{2}{3} \;
 \]
Therefore, the values we get after solving the Equation are: $1$ and $\dfrac{2}{3}$
So, the correct answer is “$\dfrac{2}{3}$ and 1”.

Note: In order to determine the factors of the above quadratic question use the Splitting up the middle or mid-term factorization method.
Note: 1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. While Using the mid-term factorization method be careful about choosing the values to split the middle term and along with this always check the sign of the midterms.