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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 +
The vector a → =(2,4) compute 3a → , ½ a → and -2a → . Graph all four vectors on similar axis system. Solution: Now here are the three scalar Multiplication 3a → = (6,
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a math problem that involves the numbers $112 for 8 hours
3 3/7 + 2 8/9 * 4=
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If roots of (x-p)(x-q) = c are a and b what will be the roots of (x-a)(x-b) = -c please explain? Ans) (x-p)(x-q)=c x2-(p+q)x-c=0 hence, a+b=p+q and a.b=pq-c
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