find the least value of k for which the equation x^2-2kx+k^2=3+x has real roots
the discriminant must be non-negative. So,
x^2-2kx+k^2 = 3+x
x^2 - (2k+1)x + (k^2-3) = 0
the discriminant is
(2k+1)^2 - 4(1)(k^2-3)
= 4k^2+4k+1 - 4k^2+12
= 4k-11
so, 4k-11 >= 0
k >= 11/4
for k= 11/4, we have
x^2 - 11/2 x + 121/16 = 3 + x
x^2 - 13/2 x + 169/16 = 0
(x - 13/4)^2 = 0
So, at k = 11/4, we have two equal roots. Any smaller value of k will cause the discriminant to be negative, producing complex roots. You can prove that for yourself by letting
k = 11/4 - h
for some positive value of h.
i believe it should13/4
as 4k+1+12>=0
4k+13>=0
k>=13/4
correct me if I'm wrong
To find the least value of k for which the equation has real roots, we need to determine the discriminant of the equation.
The discriminant (denoted by Δ) of a quadratic equation ax^2 + bx + c = 0 is given by the formula:
Δ = b^2 - 4ac.
In our case, the given equation is:
x^2 - 2kx + k^2 = 3 + x.
We can rewrite it in standard quadratic form:
x^2 - (2k + 1)x + (k^2 - 3) = 0.
Comparing this equation to the general quadratic equation ax^2 + bx + c = 0, we can see that:
a = 1,
b = -(2k + 1),
c = k^2 - 3.
Next, we substitute these values into the discriminant formula:
Δ = (-2k - 1)^2 - 4(1)(k^2 - 3).
Expanding and simplifying:
Δ = 4k^2 + 4k + 1 - 4k^2 + 12.
Δ = 4k + 13.
For the equation to have real roots, the discriminant must be greater than or equal to zero:
Δ ≥ 0.
Solving the inequality:
4k + 13 ≥ 0.
4k ≥ -13.
k ≥ -13/4.
Therefore, the least value of k is -13/4 or -3.25.
To find the least value of k for which the given equation has real roots, we need to determine the conditions that result in real roots.
Let's start by rewriting the given equation:
x^2 - 2kx + k^2 = 3 + x
Next, we can simplify the equation by rearranging it:
x^2 - x - 2kx + k^2 - 3 = 0
Now, we have a quadratic equation in standard form: ax^2 + bx + c = 0, where a = 1, b = -1 - 2k, and c = k^2 - 3.
For a quadratic equation to have real roots, the discriminant (b^2 - 4ac) must be greater than or equal to zero.
Substituting the values into the discriminant formula:
(b^2 - 4ac) ≥ 0
((-1 - 2k)^2) - 4(1)(k^2 - 3) ≥ 0
(1 + 4k + 4k^2) - 4(k^2 - 3) ≥ 0
1 + 4k + 4k^2 - 4k^2 + 12 ≥ 0
1 + 4k + 12 ≥ 0
4k + 13 ≥ 0
4k ≥ -13
k ≥ -13/4
Since we are looking for the least value of k, we need to round up the value of k to the nearest whole number. Therefore, the least value of k for which the equation has real roots is k = -3.
Confirming the solution:
Plug k = -3 back into the original equation:
x^2 - 2(-3)x + (-3)^2 = 3 + x
x^2 + 6x + 9 = 3 + x
x^2 + 5x + 6 = 0
This quadratic equation has real roots. Therefore, the least value of k for which the equation has real roots is k = -3.