# Solve $\sqrt{x}$ $+$ $\sqrt{x-\sqrt{1-x}} = 1$ and find $x$ value

It is an algebraic equation in terms of $x$ but the equation is in complicated radical form. Mathematically, it is possible to solve this equation only when the variable $x$ is brought out from the radical form.

### Square both sides of the equation

In order to free from this equation from radical form, move $\sqrt{x}$ term to right hand side and then square both sides of the equation.

$\implies$ $\sqrt{x-\sqrt{1-x}}$ $\,=\,$ $1-\sqrt{x}$

$\implies$ ${(\sqrt{x-\sqrt{1-x}})}^2$ $\,=\,$ ${(1-\sqrt{x})}^2$

$\implies$ $x-\sqrt{1-x}$ $\,=\,$ ${(1-\sqrt{x})}^2$

### Expand whole square of binomial

The right hand side expression represents square of subtraction of two terms. So, expand it by using square of subtraction of two terms formula.

$\implies$ $x-\sqrt{1-x}$ $\,=\,$ ${(1)}^2+{(\sqrt{x})}^2-2 \times 1 \times \sqrt{x}$

$\implies$ $x-\sqrt{1-x}$ $\,=\,$ $1+x-2\sqrt{x}$

Now, simplify the entire algebraic equation.

$\implies$ $x-x-\sqrt{1-x}$ $\,=\,$ $1-2\sqrt{x}$

$\implies$ $\require{cancel} \cancel{x}-\cancel{x}-\sqrt{1-x}$ $\,=\,$ $1-2\sqrt{x}$

$\implies$ $-\sqrt{1-x}$ $\,=\,$ $1-2\sqrt{x}$

### Square both sides of the equation

The algebraic equation is not completed free from radical form. So, square both sides of the equation one more time.

$\implies$ ${(-\sqrt{1-x})}^2$ $\,=\,$ ${(1-2\sqrt{x})}^2$

Now, repeat the same procedure to the right hand side expression and then simplify it further.

$\implies$ $1-x$ $\,=\,$ ${(1)}^2+{(2\sqrt{x})}^2-2 \times 1 \times 2\sqrt{x}$

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

$\implies$ $1-1-x$ $\,=\,$ $4x-4\sqrt{x}$

$\implies$ $\require{cancel} \cancel{1}-\cancel{1}-x$ $\,=\,$ $4x-4\sqrt{x}$

$\implies$ $-x$ $\,=\,$ $4x-4\sqrt{x}$

$\implies$ $4\sqrt{x}$ $\,=\,$ $4x+x$

$\implies$ $4\sqrt{x}$ $\,=\,$ $5x$

### Square both sides of the equation

The variable x is not yet completely eliminated from square root form but it can be done this time by taking square both sides of the equation.

$\implies$ ${(4\sqrt{x})}^2$ $\,=\,$ ${(5x)}^2$

$\implies$ $16x \,=\, 25x^2$

$\implies$ $25x^2-16x \,=\, 0$

The algebraic equation is completed free from radical form and transformed as a quadratic equation. Now, solve the quadratic equation for evaluating the variable $x$.

$\implies$ $x(25x-16) \,=\, 0$

Therefore, $x = 0$ or $25x-16 = 0$ is the solution of this equation.

$x = 0$ or $x = \dfrac{16}{25}$

### Verify the roots

It is time to verify the roots by substituting them in the given algebraic equation one after one.

###### 1. Put $x = \small 0$

$\sqrt{x}$ $+$ $\sqrt{x-\sqrt{1-x}}$ $\,=\,$ $\sqrt{(0)}$ $+$ $\sqrt{(0)-\sqrt{1-(0)}}$

$=\,\,\,$ $0$ $+$ $\sqrt{0-\sqrt{1}}$

$=\,\,\,$ $\sqrt{0-1}$

$=\,\,\,$ $\sqrt{-1}$

The value of the algebraic expression $\sqrt{x}$ $+$ $\sqrt{x-\sqrt{1-x}}$ is $\sqrt{-1}$ for $x$ is equal to zero but it should be one as per the given equation. So, the value $x$ should not be equal to $0$ and it cannot be a root of the given equation.

###### 2. Put $x = \small \dfrac{16}{25}$

$\sqrt{x}$ $+$ $\sqrt{x-\sqrt{1-x}}$ $\,=\,$ $\sqrt{\Bigg(\dfrac{16}{25}\Bigg)}$ $+$ $\sqrt{\Bigg(\dfrac{16}{25}\Bigg)-\sqrt{1-\Bigg(\dfrac{16}{25}\Bigg)}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\Bigg(\dfrac{16}{25}\Bigg)-\sqrt{\dfrac{25-16}{25}}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\dfrac{16}{25}-\sqrt{\dfrac{9}{25}}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\dfrac{16}{25}-\dfrac{3}{5}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\dfrac{16 \times 1 \,-\, 3 \times 5}{25}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\dfrac{16-15}{25}}$

$=\,\,\,$ $\dfrac{4}{5}$ $+$ $\sqrt{\dfrac{1}{25}}$

$=\,\,\,$ $\dfrac{4}{5}+\dfrac{1}{5}$

$=\,\,\, \dfrac{4+1}{5}$

$=\,\,\, \dfrac{5}{5}$

$=\,\,\, \require{cancel} \dfrac{\cancel{5}}{\cancel{5}}$

$=\,\,\, 1$

When $x$ is equal to $\dfrac{16}{25}$, the value of $\sqrt{x}$ $+$ $\sqrt{x-\sqrt{1-x}}$ is equal to $1$. It is the given algebraic equation. Therefore, $x = \dfrac{16}{25}$ is only one root of the equation.

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