5 Quadratic Equations. Multiple Choice. Best answer recieves 10 points.?

Question

All you have to do is match the terms with the definitions. Thanks everyone!

1.y = x^2 -5x + 4

2. y = 4x^2 + 12x + 9

3. y = 2x^2 - 10x - 1

4. y = x^2 + x + 1

Answer Choices:

A) 1 rational x-intercept

B) 0 x-intercepts

C) 2 irrational x-intercepts

D) 2 rational x-intercepts

Answer 1

Solve this by evaluating the Discriminant. That's the

b^2-4ac term in the Quadratic Formula

1. y=x^2-5x+4

a=1, b=-5, c=4

b^2-4ac=25-19, =9

Positive, and a perfect square therefore 2 rational x-intercepts. Solution is D

2.y=4x^2+12x+9

b^2-4ac=144-144,=0

Answer is A

3.y=2x^2-10x-1

b^2-4ac=100+8, =108

root 108 will be rt(36)rt(3), =6rt(3)

Therefore C

y=x^2+x+1

b^2-4ac=1-4, = -3

negative, therefore B

Answer 2

1.D

2.A

3.C

4.B

Answer 3

1) y= (x-4)(x-1) = 2 rational x-intercepts

2) y= (2x+3)^2 = 1 rational x-intercept

3) y= (10+-sqrt(100+8)) / 4 = 2 irrational x-intercepts

4) y = (-1+-sqrt(1-4))/ 2 = 0 x-intercepts

1) D

2) A

3) C

4) B

Answer 4

1 D

2 A

3 C

4 B

- - - - - - - - - - - - - - - - - - - -

1.y = x^2 -5x + 4

y = (x-4) (x -1)        .......  x-inertrcepts 1, 4

2. y = 4x^2 + 12x + 9

y = (2x + 3)*(2x + 3)  .... x intercept  -3/2

3. y = 2x^2 - 10x - 1

y = (10 + sqrt(100 + 8)) / 4  or (10 - sqrt(100 + 8)) / 4  

... irrationals since sqrt(108) is irrational

4. y = x^2 + x + 1

cyclotonic polynmial

y = (-1 + sqrt(1-4))/2 or (-1 - sqrt(1-4))/2

sqrt(-3) not real number => no x-intercepts

Answer 5

b^2 - 4ac = discriminant

for 1. b = -5 c = 4 a = 1

25 -4*1*4

= 9 so it is irrational.

9 is a perfect square (square root of it is an integer)

so

1 = D

2. a = 4, b = 12, c = 9

144 - 4*4*9

= 0

so 1 rational x intercept

2 = A

3. a = 2, b = -10 c = -1

100 - 4*2*-1

= 108

Not a perfect square

3 = C

4. a = 1 b = 1 c =1

1 - 4*1*1

= -3

negative therefore has no roots (or imaginary roots).

therefore B.

So

1=D

2=A

3=C

4 = B

Answer 6

as we know, ax^2+bx+c=0 has roots:

(-b+((b^2-4ac)^1/2))/2a and

(-b-((b^2-4ac)^1/2))/2a

1 rational x-intercept means that f(x)=0 only have 1 rational roots or (b^2-4ac)=0 so that it will only make 1 result in x when f(x)=0

0 x intercepts means it don't touch the x-axis or (b^2-4ac) is less than 0, which make the square root of it becomes imaginer number.

2 irrational roots means that (b^2-4ac) is not a square number, so when it is square root, its result is not integer.

2 rational x-intercepts means that (b^2-4ac) is a square number and when f(x)=0, there are 2possibilities of value of x.

1. x^2-5x+4=y. When y =0: b^2-4ac=25-16=9 which is a square number. So it has 2 rational x-intercepts, which is 4 and 1. The other method is by factorize it to (x-4)(x-1)=0 --> x=4 or x=1

2. 4x^2 +12 x +9=y when y=0: b^2-4ac=144-144=0 So it only has 1 rational x-intercepts, which is -3/2. Other method is by factorizing it into (2x+3)^2=0-->x=-3/2

3. 2x^2-10x-1=y. When y=0 : b^2-4ac=100+8=108 which is not a square number. So it has 2 irrational x-intercepts, which is (-10+(108^1/2))/4 and (-10-(108^1/2))/4.

4. x^2+x+1=y. When y=0, b^2-4ac=-1-4=-5 which is less than 0 and will make an imaginer number. This means that it doesn't intercepts x-axis in real number.

So:

1-->d

2-->a

3-->c

4-->b

Answer 7

The solution for x^2 -5x + 4 = 0 is x = 4,1, so it has 2 rational x intercepts.

The solution for 4x^2 + 12x + 9  = 0 is x = -1.5,-1.5, so it has 1 rational x intercept.

The solution for 2x^2 - 10x - 1  = 0 is x = 5.0981,-.0981 (both rounded off irrational nos.), so it has 2 irrational x intercepts.

The solution for x^2 + x + 1   = 0 is x = -.5+8i,-.5-.8i, so it has 0 x intercepts (not real)