Question

Solve the following equation:
$3{x^2} + 15x + 2\sqrt {{x^2} + 5x + 1} = 2$

Answer 1

First of all take the given expression, take out the common multiples common from the expression and simplify the equation assuming the bigger term by taking some reference variable. Find the value of the assumed variable and then replace it with the original term.

Complete answer:
Take the given expression: $3{x^2} + 15x + 2\sqrt {{x^2} + 5x + 1} = 2$
Take common multiples between the first two terms –
$3({x^2} + 5x) + 2\sqrt {{x^2} + 5x + 1} = 2$
Now, suppose ${x^2} + 5x = y$ …. (A)
$3(y) + 2\sqrt {y + 1} = 2$
Take term from the right hand side of the equation to the left and the root term from the left side to the right hand 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 becomes negative and vice-versa.
$3y - 2 = - 2\sqrt {y + 1}$
Take square on both the sides of the equation –
${\left( {3y - 2} \right)^2} = {\left( { - 2\sqrt {y + 1} } \right)^2}$
Square and square root cancel each other on the right hand side of the equation. And apply identity for the square of the binomial on the left hand side of the equation.
$9{y^2} - 12y + 4 = 4(y + 1)$
Multiply the term outside the bracket with the terms inside the bracket on the right hand side of the equation.
$9{y^2} - 12y + 4 = 4y + 4$
Move all the terms on one side of the equation –
$9{y^2} - 12y - 4y + 4 - 4 = 0$
Like terms with the same value cancels each other.
$9{y^2} - 16y = 0$
Take common multiple common –
$y(9y - 16) = 0$
There are two cases –
$y = 0$and II) $y = \dfrac{{16}}{9}$
By using the value of equation (A)
${x^2} + 5x = 0$
Take common multiple common from the above expression
$x(x + 5) = 0$
Make the required term the subject, when you move any term from one side to another then the sign of the terms also changes-
$\Rightarrow x = 0, - 5$ or
${x^2} + 5x = \dfrac{{16}}{9}$
Perform cross multiplication -
$9{x^2} + 45x = 16$
Move all terms on one side of the equation –
$9{x^2} + 45x - 16 = 0$
Split middle term in such a way that its product is equal to the product of first and the last term –
$\underline {9{x^2} + 48x} - \underline {3x - 16} = 0$
Find common factor from the paired terms –
$3x(3x + 16) - 1(3x + 16) = 0$
Take common term common from the above expression –
$(3x + 16)(3x - 1) = 0$
Make required term the subject –
$\Rightarrow x = - \dfrac{{16}}{3},\dfrac{1}{3}$
Hence, the value for $x = 0, - 5, - \dfrac{{16}}{3},\dfrac{1}{3}$

Note:
Be careful about the sign convention while moving any term from one side to the other and remember when you move any term from one side to another then the sign of the terms also changes. Positive term becomes negative and vice-versa. Here we received four values for “x” since the given equation was in the form of square and square-root and the most simplified form becomes the “x” to the power four. Be careful about the number of roots.