Question

$\sqrt{x + 4 } = x - 1$
​

Answer 1

Answer:

$\large\boxed{\sf{x = \frac{3 \pm \sqrt{21} }{2} }}$

Step-by-step explanation:

Given an equation such that,

$\sqrt{x + 4} = x - 1$

On squarring both the sides, we get,

$= > {( \sqrt{x + 4} )}^{2} = {(x - 1)}^{2}$

Solving further, we will get,

$= > x + 4 = {x}^{2} - 2x + 1 \\ \\ = > {x}^{2} - 2x - x + 1 - 4 = 0 \\ \\ = > {x}^{2} - 3x - 3 = 0$

Clearly, it's a quadratic equation.

Now, to find the value of x, we have,

$= > x = \dfrac{ - ( - 3) \pm \sqrt{ {( - 3) }^{2} - 4(1)( - 3)} }{2(1)} \\ \\ = > x = \dfrac{3 \pm \sqrt{9 + 12} }{2} \\ \\ = > x = \frac{3 \pm \sqrt{21} }{2}$

Hence, the values of $\bold{ x = \frac{3 \pm \sqrt{21} }{2} }$

Answer 2

Given: $\tt \sqrt{x\ +\ 4}\ =\ x\ -\ 1$

To find: The value of x.

Answer:

Let's square both the sides of the equation.

$\tt \bigg(\sqrt{x\ +\ 4}\bigg)^2\ =\ \bigg(x\ -\ 1\bigg)^2\\\\\\x\ +\ 4\ =\ x^2\ -\ 2x\ +\ 1$

Sorting out all the like terms and solving,

$\tt x\ +\ 2x\ +\ 4\ -\ 1\ =\ x^2\\\\\\3x\ +\ 3\ =\ x^2\\\\\\x^2\ -\ 3x\ -\ 3\ =\ 0$

We now have a quadratic equation.

Using the quadratic formula to solve it,

[FORMULA]:

$\tt x\ =\ \dfrac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}$

From the equation,

• a = 1
• b = -3
• c = -3

Using them in the formula,

$\tt x\ =\ \dfrac{-(-3)\ \pm\ \sqrt{(-3)^2\ -\ 4\ \times\ 1\ \times\ (-3)}}{2\ \times\ 1}\\\\\\x\ =\ \dfrac{3\ \pm\ \sqrt{9\ +\ 12}}{2}\\\\\\\bf x\ =\ \dfrac{3\ \pm\ \sqrt{21}}{2}$