Need help with a quadratic worksheet?

Question

first of all im not asking you to do my homework. Im doing IB High Level Math and am struggling so i asked the teacher for some practice work. if you could show your work so i can understand the steps to solve the problems that would be great

1. Find the range of values for p such that 3x^2+3px+p^2=1 has real roots.

2. Find the exact solutions to the equation √(3x-2) - √(x-2)=1

3.solve the inequality (x^2-2x-10)/(x+1) >= 2

Answer 1

for aan equation to have real roots the DELTA should not be less than zero so Delta = (3p)^2 - 4 *(3)*(p^2 -1) >= 0 then
9 * p^2 -12* p^2 +12 >= 0 and p^2 <= 4 and the range of p is :
[-2,2]

Answer 2

1. A quadratic equation ax²+bx+c=0 has real roots when the discriminant b²-4ac>=0. Write your equation in the correct form:

3x²+3px+(p²-1)=0

Notice that you must have 0 on the right. Plug a, b, and c into the discriminant formula and solve for p:

(3p)²-4*3(p²-1)>=0
9p²-12p²+12>=0
12-3p²>=0
p²<=4
-2<=p<=2

Your range is [-2, 2].

2. Put a radical on each side, then square the equation. Be sure to expand it properly:

√(3x-2)=√(x-2)+1
(√(3x-2))²=(√(x-2)+1)²
3x-2=(x-2)+2√(x-2)+1

Isolate the radical term:

2√(x-2)=2x-1

Square it again:

4(x-2)=4x²-4x+1
4x-8=4x²-4x+1
4x²-8x+9=0
x=(8±√(64-144))/8
x=1±√-80/8

This has no real roots, but the complex ones are x=1+4i√5/8 and x=1-4i√5/8. They need to be checked in the original equation, because raising an equation to a power often produces extraneous roots, but I'm not going to get into that here.

3. x²-2x-10>=2(x+1)
x²-2x-10>=2x+2
x²-4x-12>=0
(x+2)(x-6)>=0

A product is positive when both factors are the same sign, so separate the inequality into two parts:

(x+2>=0 AND x-6>=0) OR (x+2<=0 AND x-6<=0)
(x>=-2 AND x>=6) OR (x<=-2 AND x<=6)

All x that are greater than 6 are also greater than -2, and all x that are less than -2 are also less than 6:

x>=6 OR x<=-2

Answer 3

Just remember important thing in quadratic.
For roots to be real, D = Discriminant > 0
If D = 0, then roots repeated.
If D < 0, then complex roots.

In solving any quadratic eqn, try to easily convert the eqn in the general form,
ax^2 + bx + c = 0 where a,b,c are as simple and small as possible. So, you can easily use

x = [-b +- sqrt (D) ]/2a
If you cannot use this try factorizing the eqn.

for eg,
√(3x -2) = 1 + √(x-2)
3x - 2 = 1 + x - 2 + 2√(x-2)
2x-1 = 2√(x-2)
4x^2 - 4x + 1 = 4x - 8
4x^2 -8x + 9 = 0

The roots of the eqn are complex.

Answer 4

1. Use quadratic equation. The real roots are the values that make the quantity under the radical 0 or positive.

so basically solve for b^2 - 4ac >= 0 for p.

2. Add sqrt (x-2) to both sides of the eqn. (cancelling it out on the left side). Square both sides of the equation. solve for x.

3. Treat the inequality like an = sign.

Answer 5

1. 3x^2+3px+p^2-1>/=0
b^2-4ac>/=0
9p^2-4(3)(p^2-1)>=0
9p^2-12p^2+12>=0
-3p^2>=-12
p^2>=4
p>=-2 and p>=2

2.(√(3x-2) - √(x-2)=1)^2
(3x-1)-(x-2)=1
2x-4=0
x=-2

3.(x^2-2x-10) >= 2x+2
x^2-4x-12 >= 0
(x-6)(x+2) >= 0
x >= 6 or x >= -2

I hope this helps. :)

Answer 6

Question 1: p=1 3(x)(x)
That's all I can give now, got to go