Math Doubts

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.

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

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

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

Solve the quadratic equation

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.

Math Doubts
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more