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

[@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.

[@x81shreyasdhoke93]

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?

[@Taric25]

Yes

[@ll-lw8vn]

it would just be k>-13/4 instead k>= -13/4

[@anonymouscheesepie3768]

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

[@marksandsmith6778]

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

[@Nothingx303]

An ordinary candy 🍬 flavor math

[@m.h.6470]

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

[@creamyscroll2485]

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

[@milesrahn6093]

most don’t know how to set up the problem or what exactly they need to do. when someone tells you step-by-step what you need to do in a KZbin video, of course you’re gonna look at it and think it’s easy.

[@sidharthpatra4751]

​@@milesrahn6093This problem is very easy for anyone who knows solving quadratic equations either using formula or completing squares. That person will directly check discriminant and solve the problem. But for one who doesn't know solving roots , it will take time