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In polynomials you have seen expressions of the form x2 + 3x - 4. Also we know that when an expression is equated to zero or some other expression, we call it an equation. The equations of the second degree in a single variable "x" or "y" are generally referred to as quadratic equations and the most general form of it is
ax2 + bx + c = 0. The roots or solution for the quadratic equation can be obtained by substituting different values for x and selecting that value for which the value of the equation is zero. The methods which we have seen in factorization of polynomials are also applicable to obtain the roots of a quadratic equation. However, in this part we look at a specific method which is only applicable to solve quadratic equations.
According to this method the roots of a quadratic equation ax2 + bx + c = 0 are
x
This is derived as follows. We have
ax2 + bx + c = 0
ax2 + bx = - c ........(1)
In order to make the LHS a perfect square, we add to to the LHS and since the equality is to be preserved we do so for the other side also. Hence we obtain
x2 +
R is called as a transitive relation if (a, b) € R, (b, c) € R → (a, c) € R In other terms if a belongs to b, b belongs to c, then a belongs to c. Transitivity be uns
a²+b²=1 a+b
#quwhat is4 5/7 of 2/3estion..
Standard Basis Vectors The vector that is, i = (1, 0,0) is called a standard basis vector. In three dimensional (3D) space there are three standard basis vectors, i → = (1
What is the LCM of 4, 6, 18
A,B,C are natural numbers and are in arithmetic progressions and a+b+c=21.then find the possible values for a,b,c Solution) a+b+c=21 a+c=2b 3b=21 b=7 a can be 1,2,3,4,5,6 c c
1. Using suffix trees, give an algorithm to find a longest common substring shared among three input strings: s 1 of length n 1 , s 2 of length n 2 and s 3 of length n 3 .
Q. How to calculate Percentiles? Ans. In a large group of standardized test scores we expect the scores to approximate a normal curve. If all scores are translated to z-s
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Differentiate the following functions. (a) f (t ) = 4 cos -1 (t ) -10 tan -1 (t ) (b) y = √z sin -1 ( z ) Solution (a) Not much to carry out with this one other
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