Solve the equation x^2+(x-1)^2=3

x^2+\left(x-1\right)^2=3

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

$x_1=1.8229,\:x_2=-0.8229$

Step by step solution

Problem

$x^2+\left(x-1\right)^2=3$
1

Expanding the polynomial

$x^2+1-2x+x^2=3$
2

Adding $x^2$ and $x^2$

$1-2x+2x^2=3$
3

Rewrite the equation

$-3-2x+2x^2=0$
4

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=2$, $b=-2$ and $c=-3$

$x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
5

Substituting the values of the coefficients of the equation in the quadratic formula

$x=\frac{-2\left(-1\right)\pm \sqrt{{\left(-2\right)}^2-3\cdot 2\left(-4\right)}}{2\cdot 2}$
6

Multiply $-1$ times $-2$

$x=\frac{2\pm \sqrt{24+{\left(-2\right)}^2}}{4}$
7

Calculate the power

$x=\frac{2\pm \sqrt{24+4}}{4}$
8

Add the values $4$ and $24$

$x=\frac{2\pm \sqrt{28}}{4}$
9

Calculate the power

$x=\frac{2\pm \sqrt{28}}{4}$
10

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

$x_1=\frac{2+ \sqrt{28}}{4}\:\:,\:\:x_2=\frac{2- \sqrt{28}}{4}$
11

Simplifying

$x_1=1.8229,\:x_2=-0.8229$
12

We found that the two real solutions of the equation are

$x_1=1.8229,\:x_2=-0.8229$

Answer

$x_1=1.8229,\:x_2=-0.8229$

Problem Analysis

Main topic:

Quadratic formula

Time to solve it:

0.24 seconds

Views:

156