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Solve the quadratic equation $x^2+\left(x-1\right)^2=3$

Step-by-step Solution

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Final answer to the problem

$x=\frac{1}{2}+\sqrt{\left(\frac{5}{4}\right)},\:x=\frac{1}{2}-\sqrt{\left(\frac{5}{4}\right)}$
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Step-by-step Solution

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1

A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$

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

Combining like terms $x^2$ and $x^2$

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

Group the terms of the equation by moving the terms that have the variable $x$ to the left side, and those that do not have it to the right side

$2x^2-2x=3-1$
4

Subtract the values $3$ and $-1$

$2x^2-2x=2$
5

Rewrite the equation

$2x^2-2x-2=0$
6

Use the complete the square method to factor the trinomial of the form $ax^2+bx+c$. Take common factor $a$ ($2$) to all terms

$2\left(x^2-x-1\right)=0$
7

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$2\left(x^2-x-1+\frac{1}{4}-\frac{1}{4}\right)=0$
8

Factor the perfect square trinomial $x^2+-xx+\frac{1}{4}$

$2\left(\left(x-\frac{1}{2}\right)^2-1-\frac{1}{4}\right)=0$
9

Subtract the values $-1$ and $-\frac{1}{4}$

$2\left(-\frac{5}{4}+\left(x-\frac{1}{2}\right)^2\right)=0$
10

Multiply $-1$ times $\frac{1}{2}$

$2\left(-\frac{5}{4}+\left(x-\frac{1}{2}\right)^2\right)=0$
11

Divide both sides of the equation by $2$

$\frac{2\left(-\frac{5}{4}+\left(x-\frac{1}{2}\right)^2\right)}{2}=\frac{0}{2}$
12

Simplifying the quotients

$-\frac{5}{4}+\left(x-\frac{1}{2}\right)^2=\frac{0}{2}$
13

Divide $0$ by $2$

$-\frac{5}{4}+\left(x-\frac{1}{2}\right)^2=0$
14

We need to isolate the dependent variable , we can do that by simultaneously subtracting $-\frac{5}{4}$ from both sides of the equation

$\left(x-\frac{1}{2}\right)^2=\frac{5}{4}$
15

Removing the variable's exponent

$\sqrt{\left(x-\frac{1}{2}\right)^2}=\pm \sqrt{\left(\frac{5}{4}\right)}$
16

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

$x-\frac{1}{2}=\pm \sqrt{\left(\frac{5}{4}\right)}$
17

We need to isolate the dependent variable , we can do that by simultaneously subtracting $-\frac{1}{2}$ from both sides of the equation

$x=\pm \sqrt{\left(\frac{5}{4}\right)}+\frac{1}{2}$
18

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{\left(\frac{5}{4}\right)}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$x=\frac{1}{2}+\sqrt{\left(\frac{5}{4}\right)},\:x=\frac{1}{2}-\sqrt{\left(\frac{5}{4}\right)}$
19

Combining all solutions, the $2$ solutions of the equation are

$x=\frac{1}{2}+\sqrt{\left(\frac{5}{4}\right)},\:x=\frac{1}{2}-\sqrt{\left(\frac{5}{4}\right)}$

Final answer to the problem

$x=\frac{1}{2}+\sqrt{\left(\frac{5}{4}\right)},\:x=\frac{1}{2}-\sqrt{\left(\frac{5}{4}\right)}$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve for xFind the rootsSolve by factoringSolve by quadratic formula (general formula)Find break even pointsFind the discriminant

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Function Plot

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

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

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

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.

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