Help.. find the least k so the given quadratic equation has real solutions Reddit A-level maths

We are given x^2-2kx+k^2=3+x and have to find the least value of k so the quadratic equation has real solutions. This is a typical A-level maths problem that requires you to know about quadratic equations and discrimination. Subscribe to @bprpmathbasics

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#math #algebra #mathbasics

[ProactiveYellow]

We want the least value of k to give real roots to [equation].
1. Notice the LHS is a perfect square, so (x-k)²=x+3
2. Square rooting both sides would be nasty, so let's add and subtract k to get (x-k)²=(x-k)+(3+k)
3. Move everything over to one side to get the quadratic in (x-k) of (x-k)²-(x-k)-(3+k)=0
4. The roots are real if, and only if, the discriminant of this quadratic is non-negative, that is 1-4(1)(-3-k)≥0
5. Distributing through and simplifying we get 13+4k≥0, or 4k≥-13.
6. Fully isolating k gives us that k≥-13/4, thus to have real roots, k must be -13/4 at minimum, and to have two DISTINCT real roots, k may be any value greater than -13/4.
7. And that's a good place to stop.

[anonymouscheesepie]

x^2 - (2k + 1)x + k^2 - 3 = 0
b^2 - 4ac = 0
(2k + 1)^2 - 4(k^2 - 3) = 0
4k^2 + 4k + 1 - 4k^2 + 12 = 0
4k + 13 = 0
k = -13/4

[xshreyasdhoke]

Yo cheers!! Encountering these questions make me think of wavy curve method. Would love to see you make a video on that!! (Not for this one tho, this one's easy)

[Engy_Wuck]

what is the minimum value for two real solutions? -13/4+ε?

[marksandsmith]

Yes.
STRICT inequality for real rootS

[quigonkenny]

x² - 2kx + k² = 3 + x
(x-k)² = 3 + x
u² = 3 + (u+k) --- u = x-k
u² - u - (3+k) = 0
u² - u - m = 0 --- m = 3+k
u = [-1±√((-1)²-4(1)(-m))]/2(1)
u = -1/2 ± √(4m+1)/2

4m+1 ≥ 0
4m ≥ -1
m ≥ -1/4
(3+k) ≥ -1/4
k ≥ -1/4 - 3
k ≥ -13/4
kₘᵢₙ= -13/4

[m.h.]

Solution:



x² - 2kx + k² = x + 3 |-x -3
x² - 2kx - x + k² - 3 = 0
x² - (2k + 1)x + (k² - 3) = 0
x = ((2k + 1) ± √( (2k + 1)² - 4(k² - 3))) / 2
x = k + (1 ± √(4k² + 4k + 1 - 4k² + 12)) / 2
x = k + (1 ± √(4k + 13)) / 2
So as long as we have 4k ≥ -13 or k ≥ -13/4, we get real solutions
This is also the only point, where x has only one solution:
x = -13/4 + (1 ± √(4 * -13/4 + 13)) / 2
x = -13/4 + (1 ± √(-13 + 13)) / 2
x = -13/4 + (1 ± √0) / 2
x = -13/4 + 1/2
x = -13/4 + 2/4
x = -11/4

[creamyscroll]

meh, that was easy af, not sure why people don't just straight up annihilate the problem with the quadratic formula, makes the damn thing a cake walk really