Review question

# Can we solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}?$ Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6431

## Solution

1. Solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}.$
Squaring both sides of the equation gives us \begin{align*} &(4x+13)+(x+1)-2\sqrt{(4x+13)(x+1)} = 12-x\\ \Longrightarrow\quad& 5x+14-12+x = 2\sqrt{(4x+13)(x+1)}\\ \Longrightarrow\quad& 3x+1 = \sqrt{(4x+13)(x+1)}. \end{align*} Squaring again implies that \begin{align*} &9x^2+6x+1 = (4x+13)(x+1)=4x^2+17x+13 \\ \Longrightarrow\quad& 5x^2-11x-12 = 0 \\ \Longrightarrow\quad& (5x+4)(x-3) = 0 \end{align*}

and so $x = -\dfrac{4}{5}$ or $x = 3$.

We know now that these are the only two possibilities for a solution. There is a problem, however; squaring may have intrduced false roots.

For example, if we have $x=\sqrt{x+2}$, then squaring gives $x^2 = x+2$, and so $(x-2)(x+1)=0$, and $x = 2$ or $-1$.

But $2$ works in our original equation, while $-1$ does not. So $-1$ is a false root, and $2$ is the only true solution.

So we need to check whether or not the possible solutions above satisfy our original equation.

If $x = -\dfrac{4}{5}$, then the left-hand side is $\sqrt{4\left(-\dfrac{4}{5}\right)+13}-\sqrt{-\dfrac{4}{5} + 1} = \sqrt{\dfrac{49}{5}}-\sqrt{\dfrac{1}{5}}=\dfrac{1}{\sqrt{5}}(7-1)=\dfrac{6}{\sqrt{5}},$ while the right-hand side is $\sqrt{12+\dfrac{4}{5}}=\sqrt{\dfrac{16}{5}}=\dfrac{4}{\sqrt{5}},$ which are not equal. Thus $x=-\dfrac{4}{5}$ is not a solution.

If $x=3$, then the left-hand side is $\sqrt{4\times 3+13}-\sqrt{3 + 1}=5-2=3$ while the right-hand side is $\sqrt{12-3}=\sqrt{9}=3.$ Thus $x=3$ is the only solution to the original equation.

1. One root of the equation $3x^3+14x^2+2x-4=0$ is rational. Obtain this root and complete the solution of the equation.

Let $f(x)=3x^3+14x^2+2x-4$. Assume the given rational root is $\dfrac{a}{b}$, where the highest common factor of $a$ and $b$ is $1$ and, without loss of generality, $b>0$.

Substituting this solution into $f(x)=0$ and multiplying through by $b^3$ we find $3a^3+14a^2b+2ab^2-4b^3=0.$ Note that this means $b$ divides $3a^3$ (since it divides every other term) and so $b$ is $1$ or $3$ (since it cannot share a factor with $a$).

Now $2$ divides $14a^2b+2ab^2-4b^3$, and so $2$ divides $3a^3$, and thus $2$ divides $a$, say $a = 2c$. This gives us $6c^3 + 14c^2b + cb^3 - b^3 = 0,$ and so $c$ divides $b^3$. But $c$ and $b$ have no common factor, and so $c$ must be $\pm 1$, and $a$ must be $\pm 2$.

So the only possible rational roots for the equation are $\dfrac{2}{3}, -\dfrac{2}{3}, 2$ or $-2$.

It is easy to check that $-\dfrac{2}{3}$ is the only one to satisfy the equation. Thus $x=-\dfrac{2}{3}$ is the rational root. Factorising the cubic we find $3x^3 + 14x^2 + 2x - 4 = (3x+2)(x^2+4x-2)=0.$ So either $x = -\dfrac{2}{3}$, or $x^2+4x-2=0$ and so $x=\dfrac{-4\pm\sqrt{16-4\times(-2)}}{2}=-2\pm\sqrt{6}.$