Question

Solve $\sqrt {24 - 10x} = 3 - 4x$

Answer 1

Hint: Take the given expression, and first remove square root by taking the square of the terms. use the identity for the whole square for the difference of two terms and the required term for the subject by finding the factors of the term.

Complete step-by-step answer:
Take the given expression: $\sqrt {24 - 10x} = 3 - 4x$
Take the square on both the sides of the equation –
${\left( {\sqrt {24 - 10x} } \right)^2} = {\left( {3 - 4x} \right)^2}$
Square and square root cancel each other on the left hand side of the equation and apply the identity for the whole square of the difference of two terms.
$24 - 10x = {(3)^2} - 2(3)(4x) + {(4x)^2}$
Expand the above expression, placing the values for the square of the terms in the above equation.
$24 - 10x = 9 - 24x + 16{x^2}$
Move all the terms on one side of the equation. When you move any term from one side of the equation to the opposite side then the sign of the terms also changes. Positive term changes to negative and vice-versa.
$0 = 9 - 24x + 16{x^2} - 24 + 10x$
Combine like terms together in the above expression –
$\underline { - 24x + 10x} + 16{x^2} - \underline {24 + 9} = 0$
When you combine like terms with two different signs then you have to do subtraction and give a sign of the bigger number to the resultant value.
$16{x^2} - 14x - 15 = 0$
Split the middle term for the above expression –
$16{x^2} - \underline {24x + 10} x - 15 = 0$
Make the pair of first two terms and the last two terms in the above expression –
$\underline {16{x^2} - 24x} + \underline {10x - 15} = 0$
Take common multiple common for the paired terms in the above expression –
$8x(2x - 3) + 5(2x - 3) = 0$
Take out common term common from the above expression –
$(8x + 5)(2x - 3) = 0$
Now,
$
  8x + 5 = 0 \\
  8x = - 5 \\
  x = - \dfrac{5}{8} \;
 $

or

$
  2x - 3 = 0 \\
  2x = 3 \\
  x = \dfrac{3}{2} \;
 $
Hence, the required value is $x = - \dfrac{5}{8}$and $x = \dfrac{3}{2}$
So, the correct answer is “Option B”.

Note: Be careful to split the middle term, it should be in such a way that the product of the middle term should be equal to the product of the first and the last terms. Also, the term is multiplicative if moved to the opposite side then it goes to the denominator.