# Step-by-step Solution

## Solve the equation with radicals $2\left(\sqrt{x}\right)^2-1-\sqrt{x}=0$

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### Videos

$x=1,\:x=0.25$

## Step-by-step explanation

Problem to solve:

$2\left(\sqrt{x}\right)^2-1-\sqrt{x}=0$
1

Cancel exponents $\frac{1}{2}$ and $2$

$2x-1-\sqrt{x}=0$
2

Moving the term $-1$ to the other side of the equation with opposite sign

$2x-\sqrt{x}=1$
3

We need to isolate the dependent variable $x$, we can do that by subtracting $2x$ from both sides of the equation

$-\sqrt{x}=1-2x$
4

Multiplying both sides of the equation by $-1$

$\sqrt{x}=-1+2x$
5

Removing the variable's exponent

$x=\left(-1+2x\right)^{2}$
6

Expand $\left(-1+2x\right)^{2}$

$x=4x^2-4x+1$
7

Grouping terms

$x-4x^2=-4x+1$
8

Grouping terms

$x-4x^2-\left(-4x+1\right)=0$
9

Solve the product $-\left(-4x+1\right)$

$x-4x^2+4x-1=0$
10

Adding $4x$ and $x$

$x\left(4+1\right)-4x^2-1=0$
11

Add the values $4$ and $1$

$5x-4x^2-1=0$
12

For a simpler handling of the equation, change the sign of all terms, multiplying the entire whole by $-1$

$4x^2-5x+1=0$
13

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where in this case $a=4$, $b=-5$ and $c=1$. Then substitute the values of the coefficients of the equation in the quadratic formula:

• $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{5\pm 3}{8}$
14

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x=\frac{5+3}{8},\:x=\frac{5-3}{8}$
15

Add the values $5$ and $3$

$x=\frac{8}{8},\:x=\frac{5-3}{8}$
16

Add the values $5$ and $-3$

$x=\frac{8}{8},\:x=\frac{2}{8}$
17

Solve the equation ($1$)

$x=\frac{8}{8}$
18

Divide $8$ by $8$

$x=1$
19

Solve the equation ($2$)

$x=\frac{2}{8}$
20

Divide $2$ by $8$

$x=\frac{1}{4}$
21

The $2$ solutions of the equation are

$x=1,\:x=0.25$

$x=1,\:x=0.25$

### Problem Analysis

$2\left(\sqrt{x}\right)^2-1-\sqrt{x}=0$

### Main topic:

Equations with square roots

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