Form a quadratic equation in the form ax^2+bx+c=0 whose roots are b and twice the negative reciprocal of b.plz explain in detail
root1 = b
root2 = -2/b
so the quadratic is
(x-b)(x + 2/b) = 0
If b is a root, then (x-b) is a factor, since (x-b) = 0 if x=b.
Why is it x-b shldn't it be x+b
good answers
To form a quadratic equation with roots b and twice the negative reciprocal of b, we need to determine the values of a, b, and c in the equation ax^2 + bx + c = 0.
Let's start by finding the first root, b. The first root is given as b, so we have one value.
The second root is twice the negative reciprocal of b, which means it can be written as -2/b.
To find a quadratic equation, we know that the sum of the roots is equal to -b/a, and the product of the roots is equal to c/a.
Since we have the first root as b and the second root as -2/b, the sum of the roots is:
b + (-2/b) = (b^2 - 2)/ b (finding the common denominator)
For the sum of the roots to be -b/a, we set it equal to -b. Thus, we get the equation:
(b^2 - 2)/ b = -b
Cross-multiplying, we have:
b^2 - 2 = -b^2
Rearranging the equation, we get:
b^2 + b^2 = 2
Combining like terms, we have:
2b^2 = 2
Dividing both sides by 2, we get:
b^2 = 1
Taking the square root of both sides, we find:
b = ±1
Since we have two possible values for b, we can create two different quadratic equations:
1) For b = 1:
The quadratic equation becomes:
a(1)^2 + 1(1) + c = 0
a + 1 + c = 0
2) For b = -1:
The quadratic equation becomes:
a(-1)^2 + (-1)(-1) + c = 0
a + 1 + c = 0
Therefore, the quadratic equations in the form ax^2 + bx + c = 0, with roots b and twice the negative reciprocal of b, are:
1) For b = 1: a + 1 + c = 0
2) For b = -1: a + 1 + c = 0