How do you solve sqrt[x+5]=3x-7 and find any extraneous solutions?

Jun 30, 2017

$x = \setminus \frac{43 + \setminus \sqrt{265}}{18} \setminus \approx 3.293$

Explanation:

Simplifying the equation would involve first getting rid of the square root:
$\setminus \sqrt{x + 5} = 3 x - 7$
${\left(\setminus \sqrt{x + 5}\right)}^{2} = {\left(3 x - 7\right)}^{2}$
$x + 5 = {\left(3 x - 7\right)}^{2}$

Simplifying further, we can expand out the square on the right side of the equation:
$x + 5 = 9 {x}^{2} - 42 x + 49$

Since this is a quadratic equation, we want to put everything on one side and then find the zeros of the expression:
$0 = 9 {x}^{2} - 43 x + 44$

Using the quadratic formula (since we cannot factor the expression),
$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \setminus \frac{- \left(- 43\right) \setminus \pm \setminus \sqrt{{\left(- 43\right)}^{2} - 4 \left(9\right) \left(44\right)}}{2 \left(9\right)}$

Simplifying gives:
$x = \setminus \frac{43 \setminus \pm \setminus \sqrt{1849 - 1584}}{18}$
$x = \setminus \frac{43 \setminus \pm \setminus \sqrt{265}}{18}$

Using a calculator to find the value of this expression,
$x \setminus \approx 3.293$ or $1.485$

Plugging the first value back into the equation, we get a valid equation:
$\setminus \sqrt{3.293 + 5} = 3 \cdot 3.293 - 7$
$2.879 \setminus \approx 2.879$

However, when we plug the second value back into the equation, we have this:
$\setminus \sqrt{1.485 + 5} = 3 \cdot 1.485 - 7$
$\setminus \sqrt{6.485} = - 2.545$

Since the right side is negative, the second value that we got for $x$ is an extraneous solution. Thus, $x$ can only equal $\setminus \frac{43 + \setminus \sqrt{265}}{18} \setminus \approx 3.293$

Jun 30, 2017

all values of $x < - 5$ are extraneous.

$x = \frac{43}{18} \pm \frac{\sqrt{265}}{18} \text{ }$ as exact values

$x \approx + 3.293 \text{ }$to 3 decimal places

$x$ only has one answer if you use what is called the Principle Square Root. That is: only positive!

Explanation:

For the solution to remain within the set of numbers called Real the square root must be of a value that is not negative.

Thus all values of $x < - 5$ are extraneous.

Given:$\text{ "sqrt(x+5)=3x-7" } \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

Square both sides

$x + 5 \text{ "=" "(3x-7)^2" "=" } 9 {x}^{2} - 42 x + 49$

Subtract $\left(x + 5\right)$ from both sides

$0 = 9 {x}^{2} - 43 x + 44$

To comply with convention write as:

$9 {x}^{2} - 43 x + 44 = 0 \text{ } , \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
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Compare to $a {x}^{2} + b x d + c = 0$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case: $a = 9 \text{; "b=-43"; } c = 44$ giving:

$x = \frac{+ 43 \pm \sqrt{{\left(- 43\right)}^{2} - 4 \left(9\right) \left(44\right)}}{2 \left(9\right)}$

$x = \frac{+ 43 \pm \sqrt{265}}{18}$

$x = \frac{43}{18} \pm \frac{\sqrt{265}}{18} \text{ }$ as exact values

$x \approx + 3.293 \mathmr{and} 1.485$ to 3 decimal places

However, as per smartspot2, substituting 1.485 back into the original question proves this value to fail if you use the 'Principle Square Root'. Thus by his approach there is only 1 value that is correct:$\text{ } x = 3.293$ to 3 decimal places