How To Use the a^2 + b^2 Formula Give Steps?
Obtain the value of (a + b)
Obtain the value of ab.
Apply the formula, \(a^2 + b^2 = (a + b)^2 - 2ab\).
Overview
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The a² + b² formula helps us find the sum of the squares of two numbers. It is not a new or separate formula but comes from the expansion of the square of a sum or difference. We can write:
a² + b² = (a + b)² – 2ab
or
a² + b² = (a – b)² + 2ab
This means that if you know the value of (a + b)² or (a – b)² and the value of ab, you can easily find a² + b². This formula is helpful in simplifying algebraic expressions and solving problems related to geometry, distance, and vectors.
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The formula a² + b² is used to find the sum of the squares of two numbers or terms. It helps in solving algebraic expressions and problems in geometry or physics. The formula can be written in two different ways:
Here, a and b are any two values. You can use whichever form is easier, depending on the values given.
There’s also another formula related to a² + b², which is less commonly known but very useful. It comes from the expansion of the sum of cubes:
a³ + b³ = (a + b)(a² + b² – ab)
From this, if you know a + b and a³ + b³, you can rearrange the formula to find a² + b². This is helpful in certain types of algebra problems where cube terms are involved.
The derivation of \(a^2 + b^2\) formula is very easy. So without wasting any second, let's derive.
We all know the famous formula of square of a binomial, that is,
=> \((a + b)^2 = a^2 + b^2 + 2ab\)
=> \(a^2 + b^2 = (a + b)^2 - 2ab\)
That's all, this is the required formula of \((a^2 + b^2)\) that we have discussed in the last section. Isn't it so easy?
Alternatively, we have,
=> \((a - b)^2 = a^2 + b^2 - 2ab\)
=> \(a^2 + b^2\) = \((a - b)^2 + 2ab\)
Hence, we have proved both the formulas of \(a^2 + b^2\)
The formula a² + b² is not just used in algebra, but also plays a big role in solving problems in math and real life. Here are some common and important ways this formula is used:
Geometry and the Pythagorean Theorem
In geometry, the formula a² + b² = c² is used in the Pythagorean theorem. This tells us that in a right-angled triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides. This formula helps in finding distances, lengths, and measurements in triangles.
Example 1. Evaluate \(a^2 + b^2\) when (a + b) = 2 and ab = 1.
Solution 1. We are given with,
=> (a + b) = 2 and
=> ab = 1
Clearly, we know that,
=> \(a^2 + b^2 = (a + b)^2 - 2ab\)
=> \(a^2 + b^2 = (2)^2 - 2×1\)
=> \(a^2 + b^2 = 4 - 2 = 2\)
Hence, the value of \(a^2 + b^2\) comes out to be 2.
Example 2. Determine the value of \((a^2 + b^2)\) given that \(a^3 + b^3\) = 12, (a + b) = 2 and ab = 1
Solution 2. We are given with,
=> \(a^3 + b^3\) = 12,
=> (a + b) = 2 and
=> ab = 1
Clearly, we know that,
=> \(a^3 + b^3 = (a + b)(a^2 + b^2 - ab)\)
=> 12 = 2\((a^2 + b^2 - 1)\)
=> 6 = \((a^2 + b^2 - 1)\)
=> \(a^2 + b^2 = 7\)
Hence, the value of \(a^2 + b^2\) comes out to be 7.
Example 3. Find the value of the sum of squares of 25 and 15 and verify it using the discussed formula.
Solution 3. We need to calculate,
=> \((25)^2 + (15)^2 = 625 + 225 = 850\) —---(1)
Now, let's verify using the discussed formula. Clearly, we know that,
=> \((25)^2 + (15)^2 = (25 + 15)^2\) - 2×25×15
=> \((25)^2 + (15)^2 = (40)^2 - 750 = 1600 - 750\)
=> \((25)^2 + (15)^2 = 850\). —-(2)
Clearly, from (1) and (2), we obtained,
=> Value in (1) = value in (2)
Hence, verified.
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