Quadratic equations? 3x^2-12×+(n-5)=0 equal roots find n?

1.
3x² - 12× + (n - 5) = 0

As the equation has equal roots, the discriminant, Δ = 0
(-12)² - 4* 3 * (n - 5) = 0
144 - 12(n - 5) = 0
12(n - 5) = 144
n - 5 = 12
n = 17


====
2.
(m - 2)x² - (5 + m) x + 16 = 0

As the equation has equal roots, the discriminant, Δ = 0
[-(5 + m)]² - 4 * (m - 2) * 16 = 0
m² + 10m + 25 - 64m + 128 = 0
m² - 54m + 153 = 0
(m - 3)(m - 51) = 0
m = 3 or m = 51


====
3.
For the equation ay² + by + c = 0
y = [-b ± √(b² - 4ac)] / (2a)

(x - 1)² - 3x + 4 = 0
(x - 1)² - 3x + 3 + 1 = 0
(x - 1)² - 3(x - 1) + 1 = 0
x - 1 = [3 ± √(3² - 4*1*1)] / 2
x - 1 = (3 ± √5) / 2
x = (5 + √5) / 2 or x = (5 - √5) / 2

1

For a quadratic equation ax² + bx + c,
the sum of the roots = -b/a.
the products of the roots = c/a.
http://www.regentsprep.org/regents/math/algtrig/ATE4/natureofroots.htm
_____________________________________________

If 3x² -12×+(n-5)=0 has equal roots, R then
Sum of roots = R + R = 2R
-b/a = -(-12)/3 = 4
2R = 4
R = 2
So the roots are both 2.

The products of the roots is R*R = 2*2 = 4
c/a = (n-5)/3
(n-5)/3 = 4
n - 5 = 12
n = 17
________________________________________________

Q2 - same method

________________________________________________

Q3
(x-1)² - 3x + 4 = 0
(x² - 2x + 1) - 3x + 4 = 0
x² - 5x + 5 = 0
Then solve in the normal way.

1).
3x^2 - 12x + (n - 5) = 0
3(x - 2)^2 = 3x^2 - 12x + 12
n - 5 = 12
n = 17
2).
(m - 2) x^2 - (5 + m) x + 16 = 0 equal roots find m
As the equation has equal roots, the discriminant, Δ = 0
D
= b^2 – 4ac
= [-(5 + m)]^2 - 4(m - 2)16
= 25 + 10m + m^2 - 64m + 128
= m^2 - 54m + 153
= (m - 3)(m - 51)
m = 3
m = 51
3).
solve for x using formula (x - 1)^2 - 3x + 4 = 0
x^2 - 5x + 5 = 0
x = (5 - √5)/2
x = (5 + √5)/2

Hint:
1/ and 2/

equal roots ---> discriminant = 0

3/
ax^2+bx+c = 0---->x = (-b+/- sqrt(b^2 - 4ac))/(2a)