Quadratic Equations

Here's a few

1.Let r and s be the roots of x^2-6x+2=0 Find (r-s)^2

(r - s)^2 = r^2 - 2rs + s^2   =  r^2 + s^2 - 2rs      (1)

The sum of the roots  =   6/1 = 6

So (r+ s)^2 = r^2 + 2rs + s^2 = 36     (2)

The product of the roots = 2   = rs

So 2rs = 4

Therefore....using (2)

r^2 + 4 + s^2 = 36

r^2 + s^2 = 32

Therefore...plugging all of this into (1)....we have that

(r - s)^2  =  r^2 + s^2 - 2rs =    32 - 4   =  28

2.Find all solutions to the equation x^2+29=10x.

x^2 - 10x + 29 = 0      complete the square on x

x^2 - 10x + 25  =  -29+ 25

(x -  5)^2 = -4         take both roots

x - 5 =  ±√-4

x - 5 = ±2i 

x = 5 ± 2i

3.For what values of j does the equation (2x+7)(x-5)=-43+jx have exactly one real solution?

2x^2 - 3x - 35  =  -43 + jx      rearrange

2x^2 - (3 + j)x + 8 = 0

This will have one solution when the disriminant = 0  .......so....

(3 + j)^2 - 4(2)(8)  = 0

(3 + j)^2 - 64 = 0    

(3 + j)^2 = 64    take both roots

3 + j = ±8

3 + j = 8            or           3 + j = -8

j = 5                                   j = -11

4.What is the largest positive integer value of m such that the equation: 3x^2-mx+7=0
has no real solutions?

Thiw will occur when the discriminant is < 0      .....so....

m^2 - 4(3)(7)   < 0

m^2 - 84 < 0

m^2 <  84

m < sqrt (84)

So 

m = 9

7.

 c                  4

____   =   ______              cross-multiply

c - 5          c - 4

c (c - 4)  = 4(c - 5)

c^2 - 4c  = 4c - 20

c^2 - 8c + 20  = 0         complete the square on x

c^2 - 8c + 16  =  - 20 + 16

(c - 4)^2 =  -4        take both roots

c - 4   =  ±√-4

c - 4 = ±2i

c = 4 ±2i

5. 

\(3x^2 - 4x + 15 = 0\\ x^2 - \dfrac{4}{3} x + 5 =0\\ \therefore (b,c) = \left(\dfrac{-4}{3},5\right)\)

6.

By considering the discriminant of the equation,

\(\Delta = b^2 - 4(2)(18) = b^2 - 108\).

we know that b² - 108 must be 0 because the equation has a double root.

\(b^2 - 108 = 0\\ b = \pm \sqrt{108}\\ b= \pm6\sqrt 3\)