Can You Find The Difference Of Squares?

Relevant wiki: System of Equations - Substitution

$(A + B)^{2} = A^{2} + B^{2} + 2AB = 34 + 2 \times 15 = 64 \Longrightarrow A + B = \sqrt{64} = 8$ , where we took the positive root as $A,B \gt 0$ .

Next, $(A - B)^{2} = (A + B)^{2} - 4AB = 64 - 4 \times 15 = 4 \Longrightarrow A - B = \sqrt{4} = 2$ , where again we took the positive root, this time since $A \gt B$ .

Finally, $A^{2} - B^{2} = (A + B)(A - B) = 8 \times 2 = \boxed{16}$ .

In response to Pi Han Goh's query, given $A \gt B$ we have that $A + B = 8, A - B = 2 \Longrightarrow A = 5, B = 3$ .

Relevant wiki: System of Equations - Substitution

From $AB = 15 ,$ let $B = \frac{15}{A}$ and substitute into $A^2 + B^2 = 34 :$

$\begin{aligned} A^2 + \frac{15^2}{A^2} &= 34 \\ \frac{A^4}{A^2} + \frac{225}{A^2} &= 34 \\ \frac{A^4 + 225}{A^2} &= 34 \\ A^4 + 225 &= 34A^2 \\ A^4 - 34A^2 + 225 &= 0 \\ (A^2 - 25)(A^2 - 9) &= 0 \end{aligned}$

Looking at positive solutions only, this gives us $A = 5$ or $A = 3 .$ When $A$ takes one value $B$ takes the other, and $A > B,$ so $A = 5$ and $B = 3 .$ This gives a final answer of $5^2 - 3^2 = 25 - 9 = 16 .$

Relevant wiki: Algebraic Identities

$\begin{aligned} (A^2 + B^2)^2 - (A^2 - B^2)^2 &=& 4(AB)^2 \\ 34^2 - (A^2 - B^2)^2 &=& 4 \times 15^2 \\ (A^2 - B^2)^2 &=& 34^2 - 4\times 15^2 = 256 = 16^2 \\ A^2 - B^2 &=& 16,-16 \end{aligned}$

But since $A>B$ , then $A^2 - B^2 > 0$ , so $A^2 -B^2 = \boxed{16}$ only.

However, is there a solution for $A$ and $B$ ?

$2AB = 30$

$A^2 + B^2 + 2AB = 34 + 30 = 64 \implies (A+B)^2 = 64 \implies A+B = 8$

$A^2 + B^2 - 2AB = 34 - 30 = 4 \implies (A-B)^2 = 4 \implies A-B = 2$

$(A+B)(A-B) = A^2-B^2 = 8*2 = \boxed{16}$

Dang!! y'all are to smart, even if I still got it right the why i thought it up was eh?!

'' so what equals 15''

( 5 times 3)

5 to power of 2 =25

3 to power of 2 =9

then,25 + 9 = 34

25 - 9 = 16

yayyyyy!!!?

Squaring the equation $AB=15$ gives $A^2B^2=225$ . Now let $x=A^2$ and $y=B^2$ . Our equations become $x+y=34$ , $xy=225$ . By Vieta, $x$ and $y$ are roots of the quadratic $x^2-34x+225$ . We can factor this as $(x-9)(x-25)$ , so we see that $x=A^2=25$ and $y=B^2=9$ . Our desired quantity is $A^2-B^2=25-9=\boxed{16}.$

All numbers given in the problem, including answers, are integers. So it is reasonable to attempt to find integer solutions for $A$ and $B$ .

The product $AB=15=5\cdot3$ . Since $A>B$ , let's set $A=5, B=3$ .

Let's check the value of $A^2+B^2$ . It is indeed $5^2+3^2=25+9=34$ .

So our guess does satisfy all conditions and $A^2-B^2=5^2-3^2=25-9=\boxed{16}$ .

What i find is that ,"The sum of square of two distinct primes greater than $2$ is always an even number which follows as $\pm 2 \mod(4)$ " ie $p_1^2 + p_2^2 = e \qquad e\equiv \pm 2\mod(4).$ See note in comment section.

Since $34\equiv \pm 2 \mod(4)$ , so $A$ and $B$ must be two distinct primes less that $7$ since $7^2> 34$ . Hence, $A =5$ and $B=3$ as $A>B$ . Therefore , $A^2-B^2 =25-9=\boxed{16}$

Similar problem

$\begin{array}{rcl} A^2-B^2&=&(A-B)(A+B)\\ &=&\sqrt((A-B)^2)\cdot\sqrt((A+B)^2)\\ &=&\sqrt(A^2+B^2-2AB)\cdot\sqrt(A^2+B^2+2AB)\\ \end{array}$

Given that A^2 + B^2 equals a maximum of 34 and A is always less than B, with quick intuitive trail and error we can test numbers starting with 6.

6^2 = 36 which is > 34, implying that A can not be > 5 and hence, B < 5. Setting A = 5, we can calculate A^2 = 25.

Then we see that 34 - 25 = 9 to satisfy the equation. For B^2 = 9, then B = 3

Therefore, A^2 - B^2 = 5^2 - 3^2 = 25 - 9 = 16

Solución: 15 = 3 . 5 (5x5) + (3x3)= 34 5.3 = 15 5>3 A=5 B=3 25 - 9 = 16

AB =15, A^2 + B^2 = 34 imply A^2 + 2AB +B^2 = (A + B)^2 = 64 imply A + B = 8. A^2 - 2AB + B^2 = (A - B)^2 =4 imply A - B = 2. Then A^2 - B^2 = (A - B)(A + B) =2*8=16. Ed Gray

$(A+B)^2$ = $A^2 + B^2+ 2AB = 34 + 30 = 64$

This means: $A + B = 8 ; B = 8 - A$

So: $AB= A · (8-A) = 8A - A^2 = 15$

Then: $A^2 - 8A + 15 = 0$

♤A =5, ♧A= 3

Being A>B

A= 5 ; B = 3.

$A^2 - B^2 = 25 - 9 = 16$

The factors of 15 are 1,15 and 3,5. 25+9 = 34 25-9 = 16

15 is only 15 x 1 or 5 x3

We know 15^2 > 32 so it must be 5 and 3

We know a > b

So a = 5,b = 3

5^2-3^2

= 25 - 9

=16

$a^2+b^2=34$
$ab=15$

hence

$(a+b)^2=34+2\cdot15=64$
$(a-b)^2=34-2\cdot15=4$
$(a^2-b^2)^2=64\cdot4$

so that the answer is

$(a^2-b^2)=\sqrt{64\cdot4}=16$

DONT TRY TOO HARD

Step 1: A Square + B Square = 34

It means the squares of both A and B must be less than 34

Step 2: Find squares less than 34 They are 25, 16, 9, 4, 1

Step 3: Among these which of them adds up to give 34? 25 and 9.

Step 4: So A and B might be 5 and 3, satisfying the conditions --->> A>B as well as A,B are positive real numbers.

Step 5: A square - B square = 5 square - 3 square = 16

Using prime factorization we find that AB=15 means that either A or B has to be 5 and the other variable has to be 3. Since A>B, A=5 and B=3.

5^2 - 3^2 = 25 - 9 = 16

(The Term A^2 + B^2 = 34 isn't necessary to solve the problem)

(A+B)^2=A^2+B^2+2AB (A+B)^2=34+2*15 (A+B)^2=64 A+B=8 A=8-B AB=15 (8-B)B=15 B^2-8B+15=0 (B-5)(B-3)=0 B=3(A>B) A=5 A^2-B^2=25-9 =16

Since AB=15 and prime factorization of 15 is 3×5 or 5×3 we can compare it with A and B. So A =5 and B =3

We know that (A+B)^2= A^2+B^2+2AB Or. (A+B)^2-2AB= A^2+B^2 Therefore (A+B)^2-2AB= 34 (A+B)^2- 30= 34. (Because AB=15) (A+B)^2= 64 And. A+B= 8 We get A= 8-B Putting the value of A AB=15 B(8-B)= 15 Or B^2-5B-3B+15=0 (B-3)(B-5)=0. Therefore B= 5 or 3 As we have given that A>B so B=3 Putting the value of B in eq. AB=15 We get A=5 And so lastly we can easily calculate A^2-B^2=? 5^2-3^2= 16

All the possible answers are natural numbers, so it's a good idea to focus on them only. The only way to make AB=15 supposing A and B are naturals, and knowing that A>B, is to make A=5 and B=3. This solution also satisfies the first equality. So the answers to the problem must be 16.

For two numbers , Say we have 6 and 8

8² - 6² => 64 - 36 = 28

But on the other side we can solve it by (A - B )² = ( A + B )( A - B )

( 8 + 6 ) ( 8 - 6 ) => 14 * 2 = 28

Since A*B=15 and it can only be formed by pairs of (3,5) or (5,3) excluding (15,1) so according to question A is greater than B so A has to be 5 that means (A^2)-(B^2)= 25-9=16.

$A$ & $B$ are positive real numbers, such that $A>B$ , $A^2 + B^2 = 34$ , $AB = 15$

Factors of 15 are 1, 3, 5, 15. 15 squared is 225, which is much greater than 34. The only other options are 1, 3, and 5. $1^2 = 1$ and $34 - 1 = 33$ . 33 isn't a square number.

This leaves $5^2 + 3^2 = 34 \Longrightarrow 25 + 9 = 34$ therefore $A = 5 \quad B = 3$

Then we only need to substitute and get $25 - 9 = 16$

A²+B²=(A+B)²-2AB A²+B²+2AB=(A+B)² (A+B)²=34+30=64 (A+B)=8 If A+B=8, and A>B, and AB=15, then A must be 5, and B must be 3. So: A²-B²=(A+B)(A-B)=(5+3)(5-3)=16

AB=15 so AxB is 15. what can it be? 5x3=15. AxA is 5x5= 25. If you want 34, then it has to be B=9. BxB is 3x3=9 and this is correct. So now you know that A is 5 or 3 and B 5 or 3. A>B so 5 is A and 3 is B. AxA-BxB is simply 25-9=16.

Make A or B the subject one the last formula, then use substitution to solve for A or B in the second formula (you could use the third formula and use the second formula for the initial step, but that is unnecessarily inconvenient), and then solve the other variable, and then solve the problem (you could do it more efficiently in multiple ways but that would take too long to explain). I personally just used trial and error as this was an integer and I was too lazy to actually solve for the issue properly.

15 = 1×15 or 3×5

So if A = 15 and B = 1,

A squared + B squared = 225 + 1 = 226

However, the question states that A squared + B squared = 34.

So let's try A = 5 and B = 3,

A squared + B squared = 25 + 9 = 34.

So A squared - B squared = 25 - 9 = 16