# f(x) = x^2 - 3x + b find b velue if the function has only one root.

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f(x) = x^2 - 3x + b

If the function f(x) has only one root, then we know that delta is zero.

We know that:

delta = b^2 - 4*a*c

==> a = 1 b = -3 and c = b

==> delta = 9 - 4*1*b = 0

==> 9 - 4b = 0

==> 9 = 4b

==>** b = 9/4**

**Then the function f(x) will be:**

**f(x) = (x^2 -3x + 9/4)**

f(x) = x^2-3x+b has only one root.

We know that a quadratic equation of the form Ax^2+Bx+C = 0 has always two roots given by A

x1 = {-B+sqrt(B^2-4AC)}/2a and

x2 = {-B-sqrt(B^2-4AC)}/2a/.

Therefore to have only one root , x1 = x2, which is possible only when B^2-4AC = 0.

In the given case A= 1, B = -3 and C = b.

Therefore f(x) can have one root if and only if (-3)^2 = 4(1)(b).

Or 9 = 4b.

Or b = 9/4 = (3/2)^2.

Therefore if b = 9/4, then x^2-3x+b = 0 has one root and then f(x) = (x-3/2)^2 = 0. And the root is x = 3/2.