Square root!

Relevant wiki: Quadratic Equation

$(a - b)^2 = (c - b)^2 \implies (a - b) = \pm (c - b) \implies a = c \ \text{ OR } \ a = 2b - c.$

Take $(2-3)^{2}$ and $(4-3)^{2}$

$(2-3)^{2} = (-1)^{2} = 1$ and $(4-3)^{2} = 1^2=1$

Here 2 is not equal to 4, so this is an example for the above statement to be wrong.

If we take $(2-3)^2$ and $(2-3)^2$ ,

Now this an example for the above equation to be true.

So, the correct answer would be True, sometimes(or False).

Counterexample: $b=0, a=-c$ $(a\neq0)$

Then, $(a-b)\times(a-b)=a^2=(-c)^2=c^2$ and obviously $(c-b)\times(c-b)=c^2$

If we expand the parentheses:

$(a-b)^2=(c-b)^2 \implies a^2 -2ab + b^2 = c^2 - 2bc + b^2$

$a^2 -c^2 = 2ab -2bc$

$(a+c)(a-c)=2b(a-c) \implies a+c = 2b$

Although the statement holds for $a=c$ it also holds for $a+c=2b$ .

The easiest solution is to take a = 1, c = -1 and b = 0. Then 1 x 1 = -1 x -1 = 1

$(a-b)^2 = (c-b)^2 \\ (b-a)^2 = (c-b)^2 \implies 2b = a+c \ \text{(A.M)}$ shows that $a \neq c$ . Hence the statement we have is false . The statement will be true either $a = b$ or $b=c$ .

$(a-b)^2= (c-b)^2$ doesn't implies $a-b = c-b$ however, $|(a-b)| = |c-b|$ .

If we take b = 0, a and c may not necessarily be equal, as if a^2 = c^2, then a = c OR a = -c.

We have:

$\begin{aligned} \left ( a-b \right )^{2} &= \left ( c-b \right )^{2} \\ a^{2} - 2ab + b^{2} &= c^{2} - 2cb + b^{2} \\ a^{2} - c^{2} - 2b\left ( a-c \right ) & = 0 \\ \left ( a-c \right )\left ( a+c-2b \right ) &= 0\end{aligned}$

So, it's either $a = c$ or $a + c = 2b$ and thus the answer is $\boxed{\text{False.}}$

We have (a-b) (a-b) = (c-b) (c-b)--------eq1, This simplifies to a^2-2ab=c^2-2bc. Solving further we get the relation a+c=2b. This clearly indicates that a,b,c are in arithmetic progression. Therefore eq1 is true for any three consecutive terms in an arithmetic progression . Thus, this implies that eq1 is not only true for a=c but also for other values of a and c. For example, let the A.P. be 2,5,8 where a=2, b=5 and c=8, Now, (2-5) (2-5)=(8-5) (8-5) 9=9 Since LHS=RHS, the above equality holds true.

Let us take $a = 5 ,b = 3 ,c = 1$

$(a - b)^2 = (5 - 3)^2 = 2^2 = 4$

$(c - b)^2 = (1 - 3)^2 = (-2)^2 = 4$

Here, $a \neq c$ but $(a - b) \times (a - b) = (c - b) \times (c - b)$

If we observe carefully here the condition $a \neq c$ is satisfied only when a,b,c are in A.P .

Now put a = a, b = a + d, c = a + 2d

When we substitue the above in $(a - b) \times (a - b) = (c - b) \times (c - b)$ we get $d^2 = d^2$ .

This tells that our assumption that a,b,c are in A.P is true. So, it is obvious that $a \neq c$ .

Let A = 1, B = 0, C = -1. The result will be 1 = 1 through the formula, yet a != c.

counter example, a=1, b=2, c=3.

Not necessarily a=c because: (a-b)^2=(c-b)^2 ; (a-b)^2-(c-b)^2=0 ; [(a-b)-(c-b)][(a-b)+(c-b)]=0 ; (a-c)(a+c-2b)=0 therefore either a-c=0 or a+c-2b=0 where we get: a=c or a=2b-c.
So the answer is false.

Here is a proof by counterexample

Let $b = 0$

Our equation then simplifies to
$a^2 = c^2$

Which has $a = -c$ as a solution.

Therefore, a is not necessarily equal to c in all cases. Thus, the answer is False .

(7-3)(7-3)=(-1-3)(-1-3) and 7 $\neq$ -1

$(a-b)^2 = (c-b)^2 => (2-3)^2 = (4-3)^2 => (-1)^2 = (1)^2 => a = 2b - c$ or $(a-b)^2 = (c-b)^2 => m^2 = -m^2$ therefore, $a = c$ is not required since we can achieve m and -m with different values for a and c.

For a counterexample, simply let b = 0 and we have the classic assumption where two squares equal means their square roots are also equal. Clearly FALSE.

This is the same as the top solution I just wanted to point out that the simplified equation is ±(a - b) = ±(c - b) NOT (a - b) = ±(c - b), either side of the equation may be positive or negative since the square root must be taken of both sides.

Basically the question asks us if the $f(x) = (x-b)^2$ is an injective function or not. It can simply be checked graphically by drawing $y = (x-b)^2$ for an arbitrary value of $b$ and if any line parallel to x-axis, $y = k$ cuts the graph at two or more points it's not an injective function.

This clearly shows that $f(x) = (x-b)^2$ is not an injective function hence for the original question, if $(a-b) × (a-b) = (c-b) × (c-b)$ the statement $a = c$ may not be true.

The only "exception" is the case $(a-b) × (a-b) = (c-b) × (c-b) = 0$ , where the statement $a = c = b$ must be true

When it comes down to it, I was able to simplify to: a^2 - a = c^2 - c

I used a=3, then solved for what c could be. It could be 3 OR -2. Therefore, since one solution was found, then statement is false.

Counter example: $b = 0, a = -c \neq 0$ .

$(a-b)(a-b)=(c-b)(c-b)$ , let a = 2 and b = 3 results in an imaginary solution

(2-1)x(2-1) = (0-1)x(0-1)

If a = c or a + c = 2b then the above equation is true. Therefore it is not necessary that a must be equal to c

If a=b+x and c=b-x the statement is always true, no matter the value of x.

Quadratic equations tend to have to different roots/solutions, due to their u-shaped graph

(a - b)² = (c - b)²

• a² - 2ab + b² = c² - 2cb + b²

• a² - c² = 2ab - 2cb

• a² - c² = 2b(a - c)

• $\frac{a² - c² }{a - c}$ = 2b

• $\frac{(a + c) (a - c)}{(a - c)}$ = 2b

• a = 2b - c

(-1x2)^2=(1x2)^2

since the equation could be seen as $x^2=y^2$ any values that satisfy $a-b=-(c-b)$ or $a=2b-c$ will work.

The answer must be false on technicality, as you reduce the statement to a^2 - 2ab = c^2 - 2cb. You can use the quadratic formula to show that c will have 2 values, therefore not always equal to a.

a=1 b=2 c=3 a not equal to c

$(a-b) \times (a-b)$ is the same thing as $(a-b)^{2}$ , and as we know the square of a positive number is the same as the square of it's negative value. So $(a-b)^{2} = (c-b)^{2}$ is possible if $(a-b) = -(c-b)$ .

Example:

$(2-3) = -(4-3)$

$-1 = -(1)$

$-1^{2} = 1^{2}$

a^2 - b^2 = c^2 - b^2, a^2 = c^2, a = c or a = - c

Simply take b=0 and a=1 so b may be equal to-1 hence statement is false

Given that a minus b square is equal to C minus b square . Now here 2 cases arise : Case 1: A minus b = C minus b in this case the value of a is equal to that of c Case 2: A minus b = negative of C minus b that is a minus b = B minus C it implies that A plus c Is equal to twice of B so in this case a is not equal to c. Example : A is equal to 1 B = 2 and C = 3

Take $b=0$ , since it doesn't matter for the purposes of finding a counterexample.

Now we just have

$a^2=c^2$

But we know that square roots can have both positive and negative solutions, so taking the square root of both sides produces:

$\pm a=\pm c$

Because the signs can differ on both sides, a doesn't have to equal c -- it can also be the negative of c, and when they're squared, they become equal.

Example:

$a=3,b=0, c=-3$

$3^2 = (-3)^2$

If a is equal to 2 and b is equal to 1 then the result of 1x1 (that is the end of the a-b count) will be 1. But if c is 0 then the solution will be (0 - 1) x (0 - 1) that is (-1) x (-1), this is equal to 1 so the afirmative inst always right

if a = -2 and b = 2 then the answer to the first is 16. then C could be 6 and b is again 2 then the answer is 16

Counter example: If you let a=0 then you can show that c=2b. so a=0 b=1 c=2

If b=0 we end up with a² = c² which can be achieved by 1² = (-1)², here a =/= c therefore the above statement does not hold true.

If a^2 - 2ab + b^2 = c^2 -2cb + b^2, Then a^2 - c^2 = 2ab – 2cb, and a^2 - c^2 = 2b (a – c). So, (a + c)(a – c) = 2b (a – c), Thus a + c = 2b, that is, a = 2b - c

At this point, it is clear that the expression can't be simplified anymore. Hence, a ≠ c

Counter example is (3-4) (3-4)=(11-10) (11-10).... There are a lots of example such that a is not equal to c.

• a x - a its equal to a x a, then a is not equal to -a

If a=c then that means there would be no reason to have a c unless we were comparing natural to ordinals which in a logical sense would not be logical you could be taking absolutes but no one like those

(2-1)^2 = (0-1)^2

( a - b )^2 = ( c - b )^2

So how about using the following:

a = -2

b = 4

c = 10

Would make: ( -2 -4 )^2 = ( 10 - 4 )^2

-2 -4 = -6

-6^2= 36

10 - 4 = 6

6^2 = 36

36 = 36

It's a simple matter of - x - = +

(a)(a)+(a)(−b)+(−b)(a)+(−b)(−b)+−b2+2bc+−c2

=a2+−ab+−ab+b2+−b2+2bc+−c2

Combine Like Terms:

=a2+−ab+−ab+b2+−b2+2bc+−c2

=(a2)+(−ab+−ab)+(b2+−b2)+(2bc)+(−c2)

=a2+−2ab+2bc+−c2

Conjugates are indistinguishable for polynomials of real coefficients. Simple example is b=0 a=i c=-i

Imagine a parabola ( $x-b)^{2}$ , it is symmetric around $x=b$ so for every $d>0$ there will be two values of $x$ where ( $x-b)^{2}=d$ is satisfied

Take $a=1;b=2;c=3$ (where a!=c) Put in above equation: $(1-2)*(1-2)=(3-2)(3-2)$ which implies that: $(-1)^2 = (1)^2$ Therefore: $1=1$

let a =1, c =0, b= .5. then a -b = .5 and c -b = -.5, and (a -b)x(a -b) = .25, and (c - b)x(c - b -.5x-.5 = .25Ed Gray

• a2 - 2ab + b2 = c2 - 2bc + b2
• a2 - 2ab = c2 - 2bc
• a(a - 2b) = c(c - 2b)
• a = c(c - 2b)/(a - 2b)
• a /= c

If $(a - b)(a - b) = (c - b)(c - b)$

then $(a - b)^2 = (c - b)^2$

$(a - b) = \sqrt(c - b)^2$

$(a - b) = \pm (c - b)$

if taken as positive

$(a - b) = + (c - b)$

$a - b = c - b$

$a = c - b + b$

$a = c$

and if taken as negative

$(a - b) = - (c - b)$

$a - b = - c + b$

$a = c + b + b$

$a = 2b - c$

so value of $a$ can be $2b-c$ too. This makes the statement false.

LS (a-b) X (a-b) = (c-b) X (c-b) RS

a^2 + 2ab + b^2 = c^2 + 2cb + b^2

Subtract (b^2) and (c^2) from both sides and add (2ab) to both sides so that we get:

a^2 - 2ab + c^2 + b^2 - b^2 = c^2 - c^2 - 2cb + 2ab + b^2 - b^2

Once you cancel like terms you should end up with:

a^2 - c^2 = 2ab - 2cb

(a^2 - c^2) is a difference of squares and as such can be re written as:

(a+c) X (a-c) = 2b(a-c)

If we divide (a-c) from both sides we end up with:

a+c = 2b

If we wanted to isolate for either a or c we would end up with either (a = 2b - c) or (c = 2b - a). Since (c = 2b - a) =/= (c = a) the assertion is false.

The equation c=2b-a makes this equation true with infinite solutions.

Let's find out if this is true or false.

If (a - b) × (a - b) = (c - b) x (c - b), then we have:

$(a - b)^2$ = $(c - b)^2$ , so

$|a - b|$ = $|c - b|$

Now we can divide the problem in 2 cases:

(a - b) = (c - b) or (a - b) = (b - c)

In the first case, we conclude that a = c, so the statement would be true.

In the second case, however, we conclude that a - 2b = c, which is only true when b = 0, so the statement is false.