Product Property of Square Roots: Solving equations

Previously mentioned in the order of arithmetic operations, exponents come before multiplication and division which come before addition and subtraction (and parentheses always come before everything),

We will therefore first calculate the value of the expression inside the parentheses in the leftmost term (by calculating the values of the terms with exponents and roots inside the parentheses first) :$7:(5^2-\sqrt{16})\cdot3+\sqrt{3}\cdot\sqrt{3} = 7:(25-4)\cdot3+\sqrt{3}\cdot\sqrt{3}=7:21\cdot3+\sqrt{3}\cdot\sqrt{3}$where in the second stage we simplified the expression in parentheses,

Then for convenience we'll write the division operation in the leftmost term, where:

$7:21\cdot3+\sqrt{3}\cdot\sqrt{3} =\frac{7}{21}\cdot3+\sqrt{3}\cdot\sqrt{3}$Let's pause here for a moment and notice that we can actually write the expression in the following way:

$7\cdot\frac{1}{21}\cdot3+\sqrt{3}\cdot\sqrt{3}$And using the commutative property of multiplication we can write it as:

$7\cdot\frac{1}{21}\cdot3+\sqrt{3}\cdot\sqrt{3} = 7\cdot3\cdot\frac{1}{21}+\sqrt{3}\cdot\sqrt{3}$In other words, we can switch the order of operations - multiplying by 3 and dividing by 21 (which is represented by the fraction). If we return to writing it using division operation, it will look like this:

$7\cdot3\cdot\frac{1}{21}+\sqrt{3}\cdot\sqrt{3} =7\cdot3:21+\sqrt{3}\cdot\sqrt{3}$We could have done this at the beginning of the solution, but we must ensure that the operation (division or multiplication) before the number is well defined and that division operation can never be written first. Note that the commutative property is not well defined for division and therefore cannot be used for division, but after converting to a fraction (which is actually being multiplied) the commutative property is well defined (since it's now multiplication),

Let's return to the problem, in the last stage we got:

$\frac{7}{21}\cdot3+\sqrt{3}\cdot\sqrt{3}$ We'll continue and perform both multiplication and division in the first term, and calculate the value of the second term from the left by performing the multiplication:

$\frac{7}{21}\cdot3+\sqrt{3}\cdot\sqrt{3} = \frac{7\cdot3}{21}+(\sqrt{3})^2= \frac{21}{21}+3=1+3=4$When in the first stage we remembered that multiplication by a fraction is actually multiplication by the numerator of the fraction for the first term from the left, and for the second term from the left we remembered that multiplying a number by itself is raising the number to the power of 2. In the following stages we calculated the result of multiplication by the fraction's numerator and also remembered the definition of root for the second term (which states that root and exponent of the same order are inverse operations). Finally, we reduced the fraction (actually performed the division operation) and calculated the result of the addition operation

Therefore the correct answer is answer D.

Let's find side BC

Based on what we're given:

$\frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}$

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$\sqrt{2}x=\sqrt{x}BC$

Let's divide by square root x:

$\frac{\sqrt{2}\times x}{\sqrt{x}}=BC$

$\frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC$

Let's reduce the numerator and denominator by square root x:

$\sqrt{2}\sqrt{x}=BC$

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

$AB^2+BC^2=AC^2$

Let's substitute what we're given:

$x^2+(\sqrt{2}\sqrt{x})^2=m^2$

$x^2+2x=m^2$

We know that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's look at triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute in our known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

Finally, we will add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$