This video demonstrates how to solve a system of two square root equations by adding and subtracting the equations to create new variables (m and n), then applying the difference of squares identity to simplify the expressions, ultimately revealing that m = n, which allows solving for the original variables a and b.
Deep Dive
Prerequisite Knowledge
- No data available.
Where to go next
- No data available.
Deep Dive
Germany | A Nice Olympiad Algebra ProblemAdded:
Hello everyone. You're welcome to solve this nice square root math problem which is a square root of a + 3 then plus roo<unk> of b + 2. This is equal to 8.
Let's call this equation one. Then we have the square roo<unk> of a subtract 4 plus the roo<unk> of b - 5. This is equal to 6. Let's call this equation two. So what is the value of a and what is the value of b? So let provide a solution from here.
So the first step is to add equation one by equation 2.
So this implies that we have the square root of a + 3 then plus square root of a subtract 4 then + square roo<unk> of b + 2 then + square<unk> of b subtract 5.
This is equal to 8 + 6.
So we have square roo<unk> of a + 3 +<unk> of a subtract 4 +<unk> of b + 2 +<unk> of b - 5 this is equal to 8 + 6 which is 14.
Now the next step from here we can let square roo<unk> of a + 3 then plus square root of a subtract for b = m and we can let the square root of b + 2 + roo<unk> of b - 5 be = n.
So let's express this equation here in terms of m and n. So this means we have m + n. This is equal to 14.
Let's call this as equation three. Let's call this as equation three. Now the next step is let's subtract equation one by equation two.
So this means we have the square root of a + 3 then plus square roo<unk> of b + 2 then subtract square<unk> of a - 4 then we have +<unk> of b - 5. This is equal to 8 subtract 6.
So here we have the square root of a + 3 then plus square roo<unk> of b + 2. Now let's open the parenthesis here. This is - square<unk> of a - 4. Then - * + this becomes minus. We have -<unk> b - 5.
This is equal to 8 - 6 which is equal to which is equal to 2. So that now we have the square root of a + 3 subtract square<unk> of a - 4 then plus square roo<unk> of b + 2 subtract square roo<unk> of b - 5. This is equal to 2.
Now let's call this as equation four. Let's call this as equation four.
Now the next step is that given the difference of two square the difference of two squares that is m² - n² this can be expressed as m + n * m subtract n.
Now given that m is equal to the square roo<unk> of a + 3 then +<unk> of a - 4. This is multiplied by square<unk> of a + 3 then subtract square<unk> of a - 4.
Now this means that this is the same thing as square roo<unk> of a + 3 ra to ^ of 2 then subtract square<unk> of a - 4 raised ^ of 2.
This is equal to let's eliminate the square root sign here so that we have a + 3. Then now we also eliminate the square root sign here. So this is - a.
Then - * -4 this is + 4. So we have that a minus a simplifies. So that now we have 3 + 4 and this is equal to 7.
So we have the square root of a + 3 then plus square root of a subtract 4 multiplying by square<unk> of a + 3 subtract square<unk> of a a - 4 this is equal to 7.
But we have that the square root of a + 3 + roo<unk> of a minus 4. This is the same thing as m. This is what we have here. This is the same thing as m. So let's substitute m here. So that we have multip + 3 -<unk> a subtract 4. This is equal to 7. So let's divide on both sides by m. So that now we have the square roo<unk> of a + 3 subtract square roo<unk> of a - 4. This is equal to 7 / m. Let's call this equation 5.
Now similarly given n remember n this is the same thing as the square root of b b + 2 then we have plus square root of b minus 5.
Now applying the difference of perfect squares then we have that the square root of b + 2 subtract square<unk> of b - 5. This is supposed to give us a value of 7 / n.
Let's call this as equation 6.
Now we are saying that equation six this is obtained by applying the difference of two perfect squares identity similar to what we have done here for the case of m. So that now we have the square root of b + 2 subtract roo<unk> of b minus 5. This is the same thing as 7 of n. So this is our equation six.
Now from equation four from equation four which is the square roo<unk> of a + 3 subtract square roo<unk> of a - 4 then + square roo<unk> of b + 2 subtract square<unk> of b - 5 this is equal to this is equal to 2. Now we have that the roo<unk> of a + 3 - roo<unk> of a - 4.
This is equation 5. So which is 7 / m + now roo<unk> of b + 2 - roo<unk> of b - 5. This is equation 6 which is 7 / n and this is equal to okay 7 is common here. So let's factor out 7 so that we have 1 / m + 1 / n. This is equal to 2.
Let's divide both sides by 7. So that now 1 / m + 1 / n. This is equal to 2 / 7.
Let's call this equation 7.
Now the next step is to multiply equation equation three.
This is multiplying by equation 7.
Now equation three. Equation three is m + n. This is equal to 14. So we have m + n ultip by 1 / m + 1 / n. This is equal to 2 / 7 * 14.
So m * 1 / m this is m * 1 / m + m * 1 / n then + n * 1 / m. So we have n * 1 / m then + n * 1 / n. This is equal to 2 / 7 * 14. So let's simplify here. 14 / 7 this is 2. So we have 2 * 2 which is 4. So here we have m and m simplifies.
So we have 1 + m / n then + n / m then plus n and n simplify.
So this is + 1 and this is equal to 4.
Okay. So here we have 1 + 1 this is 2 + m / n + n / m this is equal to 4.
Now the next step is to take two on the right hand side so that we have m / n + n / m this is = 4 - 2.
So this is m / n + n / m. This is = 4 - 2 which is 2. Now we have m / n + n / m.
This is equal to 2. So we have the LCM here which is n m n m / n. This is m * m which is m 2 + n².
This is equal to 2. Now let's cross multiply from here. So that we have m² + n² this is equal to m n.
Now let's take 2 mn on the left hand side. So that we have m² + n² subtract 2 m n this is equal to this is equal to z.
Now you find that this equation here this is actually in the form of this is in the form of x² + y^ 2 - 2x y which we can express as x subtract y raised ^ of 2. So let's apply this algebraic identity here. So that we have mus n raised to ^ of 2 this is equal to zero.
Now so for the value of m - n let's introduce a square root on both sides.
So we have the square root of m - n ^ 2 this is equal to the root of 0. So here we have m - n. this is equal to to zero.
Such that now we have m is = n. m is = n. Now the next step is that if you recall if you recall from equation three which is m + n, this is equal to 14.
Since m is equal to n, we can substitute m with n. So that we have n + n. This is equal to 14.
So this is 2 n this is equal to 14. So let's divide both sides by 2. And this means that n is = 7. n is = 7. So this is to mean that m is also equal to 7. So we have the value of m and n.
Now since we have m is = 7, n is= 7. If you recall, if you recall, we have that the square root of a + 3, then plus square root of a subtract 4. This is the same thing as m.
Again, if you recall from equation five, we have the square root of a + 3 subtract roo<unk> of a - 4. This is the same thing as 7 / m.
So these are systems of two re equations. So let's sum these two system of ree equations. So we have square roo<unk> of a + 3 + roo<unk> of a + 3.
This is 2 square roo<unk> of a + 3. Then roo<unk> a - 4 + -<unk> a - 4. This simplifies. Then this is equal to m + 7 / m. Okay. So this is 2<unk> of a + 3.
This is equal to m. m is the same thing as 7. We have the value of m which is 7 + 7 / 7. So you simplify 7 and 7 here.
So this is 1. So we have 2 square<unk> of a + 3. This is equal to 7 + 1 and this is equal to 8.
The next step is to divide on both sides by 2. So that now square root of a + 3.
This is equal to 8 / 2 which is 4.
Now the next step is to square on both sides from here. So that now we eliminate the square root sign on the left hand side. So this is a + 3 and this is = 4^ 2 and this is = 16. 4² is 16. So let's take + 3 on the right hand side. So that a is equal to 16 subtract 3. We have the value of a is equal to 16 - 3 and this is equal to 13. So the value of a is equal to 13.
Now again if you recall if you recall again if you recall here we have that the square root of b + 2 plus the square root of b minus 5. This is the same thing as n.
And from equation 6 which is the square roo<unk> of b + 2 subtract the square roo<unk> of b - 5. This is equal to 7 / n. Again this a system of two equations.
So let's sum this system of two equations. Square root of b + 2 + roo<unk> of b + 2. This is 2 *<unk> b + 2. Then square roo<unk> of b - 5 + - roo<unk> of b - 5. This simplifies and this is equal to n + 7 / n.
Let's substitute the value of n so that we have 2 * roo<unk> of b + 2. This is equal to 7 + 7 / 7.
So this is to mean that we have 2 * b + 2 this is equal to 7 + 7 / 7. This is the same thing as 1.
So we have 2 *<unk> of b + 2. This is equal to 7 + 1 and this is equal to 8.
So let's divide both sides by 2. And what this implies is that the square roo<unk> of b + 2 this is equal to 8 / 2 and this is 4. So we have four here.
The next step is to square on both sides from here. So that now we eliminate the square root sign and this is b + 2. This is equal to 4^ 2 and this is equal to 16.
To solve for B, let's take plus two on the right hand side. And this means the value of B is equal to 16 subtract 2.
So we have B is equal to 16 - 2 and this is equal to 14. So we have the value of B = 14.
We have the value of A= to 13.
Now let's check if the value of a and b satisfies the equation.
Now let's verify that the value of a which is 13 and b which is 14 satisfies the equation from equation one which is the square root of a + 3 then plus square root of b + 2. This is supposed to give us a value of 8. So let's substitute a. We have the square root of 13 + 3 then + square roo<unk> of b which is 14 + 2. This should give us a value of 8.
So this is the square root of 13 + 3.
This is square roo<unk> of 16 + square<unk> of 14 + 2. This is square root of 16. This should give us a value of 8 square root of 16. This is 4 + 4. This should give us a value of 8. So 4 + 4 this is 16 which is equal to 16. So we have the left add side is equal to the right add side. And this affirms that the value of a which is 13 and b which is
Related Videos
Olympiad Mathematics | Indian | Can You Solve This One?
PhilCoolMath
650 views•2026-06-03
Escaping the Fog
LogicLemurGaming
760 views•2026-06-03
H2 Math June Holiday 2026 Intensive Revision | H2 Math Tuition by Achevas #singaporemath #h2math
AchevasTV
304 views•2026-06-01
A Brutal Radical Expression Made Easy! The Shortcut Changes Everything.
tamoshop
112 views•2026-06-02
V : jee main /advance class 11 mathematics : Binomial Theorem class-1 ( 29 may 2026 )
dcamclassesiitjeemainsadva9953
125 views•2026-05-29
Is This Pentomino Tileable?
3cycle
241 views•2026-05-30
This Sudoku Has Many Lines!!
CrackingTheCryptic
2K views•2026-05-29
Olympiad Mathematics | Indian Can You Solve This One?
PhilCoolMath
268 views•2026-06-02











