Show Find more answers or ask new questions now. Find More Answers 
Answer (Detailed Solution Below) Option 2 : 9 cm India's Super Teachers for all govt. exams Under One Roof Enroll For Free Now
Let longer segment and smaller segment be M and N respectively. Given, Area of square with M segment – area of square with N segment = 32 And, ⇒ M – N = 2 Then, ⇒ M2 – N2 = 32 ⇒ (M – N)(M + N) = 32 ⇒ M + N = 16 Solving equations, ⇒ 2M = 18 ⇒ M = 9 cm ⇒ N = 7 cm ∴ Length of greater line segment is 9 cm.India’s #1 Learning Platform Start Complete Exam Preparation
Video Lessons & PDF Notes Trusted by 2,98,84,403+ Students
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
A quadratic equation is a second degree polynomial usually in the form of f(x) = ax2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f (x). Every quadratic equation has two values of the unknown variable, usually known as the roots of the equation (α, β). What is Difference of Squares?The difference of two squares is a theorem that tells us if a quadratic equation can be written as a product of two binomials, in which one shows the difference of the square roots and the other shows the sum of the square roots. One thing to note about this theorem is that it does not apply to the SUM of squares. Difference of Squares FormulaThe difference of square formula is an algebraic form of the equation used to express the differences between two square values. A difference of square is expressed in the form: a2 – b2, where both the first and last term is perfect squares. Factoring the difference of the two squares gives: a2 – b2 = (a + b) (a – b) This is true because, (a + b) (a – b) = a2 – ab + ab – b2 = a2 – b2 How to Factor Difference of Squares?In this section, we will learn how to factorize algebraic expressions using the difference of square formula. To factor a difference of squares, the following steps are undertaken:
Let’s solve a few examples by applying these steps. Example 1 Factor 64 – x2 Solution Since we know the square of 8 is 64, then we can rewrite the expression as; Example 2 Factorize Solution Since x2−16 = (x) 2− (4)2, therefore apply the difference square formula a2 – b2 = (a + b) (a – b), where a and b in this case are x and 4 respectively. Therefore, x2 – 42 = (x + 4) (x – 4) Example 3 Factor 3a2 – 27b2 Solution Since 3 is GCF of the terms, we factor it out. Example 4 Factor x3 – 25x Since the GCF = x, factor it out; Example 5 Factor the expression (x – 2)2 – (x – 3)2 Solution In this problem a = (x – 2) and b = (x – 3) We now apply a2 – b2 = (a + b) (a – b) = [(x – 2) + (x – 3)] [(x – 2) – (x – 3)] = [x – 2 + x – 3] [x – 2 – x + 3] Combine the like terms and simplify the expressions; [x – 2 + x – 3] [x – 2 – x + 3] = > [2x – 5] [1] = [2x – 5] Example 6 Factor the expression 25(x + y)2 – 36(x – 2y)2. Solution Rewrite the expression in the form a2 – b2. 25(x + y)2 – 36(x – 2y)2 => {5(x + y)}2 – {6(x – 2y)}2 = [5(x + y) + 6(x – 2y)] [5(x + y) – 6(x – 2y)] = [5x + 5y + 6x – 12y] [5x + 5y – 6x + 12y] Collect like terms and simplify; = (11x – 7y) (17y – x). Example 7 Factor 2x2– 32. Solution Factor out the GCF; Applying the difference squares formula, we get; Example 8 Factor 9x6 – y8 Solution First, rewrite 9x6 – y8 in the form a2 – b2. 9x6 – y8 => (3x3)2 – (y4)2 Apply a2 – b2 = (a + b) (a – b) to get; = (3x3 – y4) (3x3 + y4) Example 9 Factor the expression 81a2 – (b – c)2 Solution Rewrite 81a2 – (b – c)2 as a2 – b2 = [9a + b – c] [9a – b + c ] In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity
in elementary algebra.
The proof of the factorization identity is straightforward. Starting from the left-hand side, apply the distributive law to get
By the commutative law, the middle two terms cancel:
leaving
The resulting identity is one of the most commonly used in mathematics. Among many uses, it gives a simple proof of the AM–GM inequality in two variables.
The proof holds in any commutative ring.
Conversely, if this identity holds in a ring R for all pairs of elements a and b, then R is commutative. To see this, apply the distributive law to the right-hand side of the equation and get
For this to be equal to
a
2
−
b
2
{\displaystyle a^{2}-b^{2}}
, we must have
for all pairs a, b, so R is commutative.
The difference of two squares can also be illustrated geometrically as the difference of two square areas in a plane. In the diagram, the shaded part represents the difference between the areas of the two squares, i.e.
a
2
−
b
2
{\displaystyle a^{2}-b^{2}}
. The area of the shaded part can be found by adding the areas of the two rectangles;
a
(
a
−
b
)
+
b
(
a
−
b
)
{\displaystyle a(a-b)+b(a-b)}
, which can be factorized to
(
a
+
b
)
(
a
−
b
)
{\displaystyle (a+b)(a-b)}
. Therefore,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
{\displaystyle a^{2}-b^{2}=(a+b)(a-b)}
.
Another geometric proof proceeds as follows: We start with the figure shown in the first diagram below, a large square with a smaller square removed from it. The side of the entire square is a, and the side of the small removed square is b. The area of the shaded region is
a
2
−
b
2
{\displaystyle a^{2}-b^{2}}
. A cut is made, splitting the region into two rectangular pieces, as shown in the second diagram. The larger piece, at the top, has width a and height a-b. The smaller piece, at the bottom, has width a-b and height b. Now the smaller piece can be detached, rotated, and placed to the right of the larger piece. In this new arrangement, shown in the last diagram below, the two pieces together form a rectangle, whose width is
a
+
b
{\displaystyle a+b}
and whose height is
a
−
b
{\displaystyle a-b}
. This rectangle's area is
(
a
+
b
)
(
a
−
b
)
{\displaystyle (a+b)(a-b)}
. Since this rectangle came from rearranging the original figure, it must have the same area as the original figure. Therefore,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
{\displaystyle a^{2}-b^{2}=(a+b)(a-b)}
.
The formula for the difference of two squares can be used for factoring polynomials that contain the square of a first quantity minus the square of a second quantity. For example, the polynomial
x
4
−
1
{\displaystyle x^{4}-1}
can be factored as follows:
As a second example, the first two terms of
x
2
−
y
2
+
x
−
y
{\displaystyle x^{2}-y^{2}+x-y}
can be factored as
(
x
+
y
)
(
x
−
y
)
{\displaystyle (x+y)(x-y)}
, so we have:
Moreover, this formula can also be used for simplifying expressions:
The difference of two squares is used to find the linear factors of the sum of two squares, using complex number coefficients.
For example, the complex roots of
z
2
+
4
{\displaystyle z^{2}+4}
can be found using difference of two squares:
Therefore, the linear factors are
(
z
+
2
i
)
{\displaystyle (z+2i)}
and
(
z
−
2
i
)
{\displaystyle (z-2i)}
.
Since the two factors found by this method are complex conjugates, we can use this in reverse as a method of multiplying a complex number to get a real number. This is used to get real denominators in complex fractions.[1] Rationalising denominatorsThe difference of two squares can also be used in the rationalising of irrational denominators.[2] This is a method for removing surds from expressions (or at least moving them), applying to division by some combinations involving square roots. For example: The denominator of 5 3 + 4 {\displaystyle {\dfrac {5}{{\sqrt {3}}+4}}} can be rationalised as follows: 5 3 + 4 {\displaystyle {\dfrac {5}{{\sqrt {3}}+4}}} = 5 3 + 4 × 3 − 4 3 − 4 {\displaystyle ={\dfrac {5}{{\sqrt {3}}+4}}\times {\dfrac {{\sqrt {3}}-4}{{\sqrt {3}}-4}}} = 5 ( 3 − 4 ) ( 3 + 4 ) ( 3 − 4 ) {\displaystyle ={\dfrac {5({\sqrt {3}}-4)}{({\sqrt {3}}+4)({\sqrt {3}}-4)}}} = 5 ( 3 − 4 ) 3 2 − 4 2 {\displaystyle ={\dfrac {5({\sqrt {3}}-4)}{{\sqrt {3}}^{2}-4^{2}}}} = 5 ( 3 − 4 ) 3 − 16 {\displaystyle ={\dfrac {5({\sqrt {3}}-4)}{3-16}}} = − 5 ( 3 − 4 ) 13 . {\displaystyle =-{\dfrac {5({\sqrt {3}}-4)}{13}}.}Here, the irrational denominator 3 + 4 {\displaystyle {\sqrt {3}}+4} has been rationalised to 13 {\displaystyle 13} . Mental arithmeticThe difference of two squares can also be used as an arithmetical short cut. If two numbers (whose average is a number which is easily squared) are multiplied, the difference of two squares can be used to give you the product of the original two numbers. For example: 27 × 33 = ( 30 − 3 ) ( 30 + 3 ) {\displaystyle 27\times 33=(30-3)(30+3)}Using the difference of two squares, 27 × 33 {\displaystyle 27\times 33} can be restated as a 2 − b 2 {\displaystyle a^{2}-b^{2}} which is 30 2 − 3 2 = 891 {\displaystyle 30^{2}-3^{2}=891} .Difference of two consecutive perfect squaresThe difference of two consecutive perfect squares is the sum of the two bases n and n+1. This can be seen as follows: ( n + 1 ) 2 − n 2 = ( ( n + 1 ) + n ) ( ( n + 1 ) − n ) = 2 n + 1 {\displaystyle {\begin{array}{lcl}(n+1)^{2}-n^{2}&=&((n+1)+n)((n+1)-n)\\&=&2n+1\end{array}}}Therefore, the difference of two consecutive perfect squares is an odd number. Similarly, the difference of two arbitrary perfect squares is calculated as follows: ( n + k ) 2 − n 2 = ( ( n + k ) + n ) ( ( n + k ) − n ) = k ( 2 n + k ) {\displaystyle {\begin{array}{lcl}(n+k)^{2}-n^{2}&=&((n+k)+n)((n+k)-n)\\&=&k(2n+k)\end{array}}}Therefore, the difference of two even perfect squares is a multiple of 4 and the difference of two odd perfect squares is a multiple of 8. Factorization of integersSeveral algorithms in number theory and cryptography use differences of squares to find factors of integers and detect composite numbers. A simple example is the Fermat factorization method, which considers the sequence of numbers x i := a i 2 − N {\displaystyle x_{i}:=a_{i}^{2}-N} , for a i := ⌈ N ⌉ + i {\displaystyle a_{i}:=\left\lceil {\sqrt {N}}\right\rceil +i} . If one of the x i {\displaystyle x_{i}} equals a perfect square b 2 {\displaystyle b^{2}} , then N = a i 2 − b 2 = ( a i + b ) ( a i − b ) {\displaystyle N=a_{i}^{2}-b^{2}=(a_{i}+b)(a_{i}-b)} is a (potentially non-trivial) factorization of N {\displaystyle N} . This trick can be generalized as follows. If a 2 ≡ b 2 {\displaystyle a^{2}\equiv b^{2}} mod N {\displaystyle N} and a ≢ ± b {\displaystyle a\not \equiv \pm b} mod N {\displaystyle N} , then N {\displaystyle N} is composite with non-trivial factors gcd ( a − b , N ) {\displaystyle \gcd(a-b,N)} and gcd ( a + b , N ) {\displaystyle \gcd(a+b,N)} . This forms the basis of several factorization algorithms (such as the quadratic sieve) and can be combined with the Fermat primality test to give the stronger Miller–Rabin primality test. Vectors a (purple), b (cyan) and a + b (blue) are shown with arrows The identity also holds in inner product spaces over the field of real numbers, such as for dot product of Euclidean vectors: a ⋅ a − b ⋅ b = ( a + b ) ⋅ ( a − b ) {\displaystyle {\mathbf {a} }\cdot {\mathbf {a} }-{\mathbf {b} }\cdot {\mathbf {b} }=({\mathbf {a} }+{\mathbf {b} })\cdot ({\mathbf {a} }-{\mathbf {b} })}The proof is identical. For the special case that a and b have equal norms (which means that their dot squares are equal), this demonstrates analytically the fact that two diagonals of a rhombus are perpendicular. This follows from the left side of the equation being equal to zero, requiring the right side to equal zero as well, and so the vector sum of a + b (the long diagonal of the rhombus) dotted with the vector difference a - b (the short diagonal of the rhombus) must equal zero, which indicates the diagonals are perpendicular. Difference of two nth powersVisual proof of the differences between two squares and two cubes If a and b are two elements of a commutative ring R, then a n − b n = ( a − b ) ( ∑ k = 0 n − 1 a n − 1 − k b k ) {\displaystyle a^{n}-b^{n}=\left(a-b\right)\left(\sum _{k=0}^{n-1}a^{n-1-k}b^{k}\right)} . Historically, the Babylonians used the difference of two squares to calculate multiplications. [3] For example: 93 x 87 = 90² - 3² = 8091 64 x 56 = 60² - 4² = 3584
|
Based on the problem number 3 what is the difference of the areas of two squares respectively
zusammenhängende Posts
Urheberrechte © © 2024 frojeostern Inc.
