Activity 2: Match My Real Values! Instruction: Match the quadratic equations in column A with their values in Column B by using factoring. Then, answer the questions that follow. Column A Column B
1.-4x-30=-2x² a. no real solutions
2. 3x²+27x=0 b. x=3; x = -8
3. x²+6x+9=0 c. x=0; x=-9
4.4x²+16=8x d. x= -3
5. x²+5x=24 e. x=5; x = -3

I need step by step explanation PLZZ.

⟨1.⟩

Given equation is -4x-30 = -2x²

⇛ -4x-30+2x² = 0

⇛ 2x²-4x-30 = 0

⇛ 2(x²-2x-15) = 0

⇛ x²-2x-15 = 0

It is the form of ax²+bx+c = 0

We have,

a = 1

b = -2

c = -15

We know that

The discriminant of ax²+bx+c = 0 is b²-4ac

Discriminant = (-2)²-4(1)(-15)

⇛ D = 4+60

⇛ D = 64 > 0

The given equation has two distinct and real roots.

Now,

x²-2x-15 = 0

⇛ x²+3x-5x-15 = 0

⇛ x(x+3)-5(x+3) = 0

⇛ (x+3)(x-5) = 0

⇛ x+3 = 0 or x-5 = 0

⇛ x = -3 or x = 5

The roots are -3 and 5.

⟨2.⟩

Given equation is 3x²+27x = 0

It is the form of ax²+bx+c = 0

We have,

a = 3

b = 27

c = 0

The discriminant of ax²+bx+c = 0 is b²-4ac

Discriminant = 27²-4(3)(0)

⇛ D = 729 > 0

The given equation has two distinct and real roots.

⇛3x(x+9) = 0

⇛ 3x = 0 or x+9 = 0

⇛ x = 0/3 or x = -9

⇛ x = 0 or x = -9

The roots are 0 and -9.

⟨3.⟩

Given equation is x²+6x+9 = 0

It is the form of ax²+bx+c = 0

We have,

a = 1

b = 6

c = 9

We know that

The discriminant of ax²+bx+c = 0 is b²-4ac

Discriminant = 6²-4(1)(9)

⇛ D = 36-36

⇛ D = 0

The given equation has two equal and real roots.

Now,

x²+6x+9 = 0

⇛ x²+3x+3x+9= 0

⇛ x(x+3)+3(x+3) = 0

⇛ (x+3)(x+3) = 0

⇛ x+3 = 0 or x+3 = 0

⇛ x = -3 or x = -3

The roots are -3 and -3.

⟨4.⟩

Given equation is 4x²+16 = 8x

⇛ 4x²+16-8x = 0

⇛ 4x²-8x+16 = 0

⇛ 4(x²-2x+4) = 0

⇛ x²-2x+4 = 0

It is the form of ax²+bx+c = 0

We have,

a = 1

b = -2

c = 4

We know that

The discriminant of ax²+bx+c = 0 is b²-4ac

Discriminant = (-2)²-4(1)(4)

⇛ D = 4-16

⇛ D = -12 < 0

The given equation has no real roots.

⟨5.⟩

Given equation is x² +5x = 24

⇛ x²+5x-24 = 0

It is the form of ax²+bx+c = 0

We have,

a = 1

b = 5

c = -24

We know that

The discriminant of ax²+bx+c = 0 is b²-4ac

Discriminant = (5)²-4(1)(-24)

⇛ D = 25+96

⇛ D = 124 > 0

The given equation has two distinct and real roots.

Now,

x²+5x-24 = 0

⇛ x²+8x-3x-24 = 0

⇛ x(x+8)-3(x+8) = 0

⇛ (x+8)(x-3) = 0

⇛ x+8 = 0 or x-3 = 0

⇛ x = -8 or x = 3

The roots are -8 and 3.

Answer:

is should be y = 5x

step-by-step explanation:

correct me if im wrong