Question

The roots of the quadratic equation 4y2 = 16y
are​

Answer 1

Answer:

Changes made to your input should not affect the solution:

(1): "y2" was replaced by "y^2".

Step by step solution :

STEP

1

:

Equation at the end of step 1

(22y2 - 16y) + 16 = 0

STEP

2

:

STEP

3

:

Pulling out like terms

3.1 Pull out like factors :

4y2 - 16y + 16 = 4 • (y2 - 4y + 4)

Trying to factor by splitting the middle term

3.2 Factoring y2 - 4y + 4

The first term is, y2 its coefficient is 1 .

The middle term is, -4y its coefficient is -4 .

The last term, "the constant", is +4

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -4 .

-4 + -1 = -5

-2 + -2 = -4 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and -2

y2 - 2y - 2y - 4

Step-4 : Add up the first 2 terms, pulling out like factors :

y • (y-2)

Add up the last 2 terms, pulling out common factors :

2 • (y-2)

Step-5 : Add up the four terms of step 4 :

(y-2) • (y-2)

Which is the desired factorization

Multiplying Exponential Expressions:

3.3 Multiply (y-2) by (y-2)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (y-2) and the exponents are :

1 , as (y-2) is the same number as (y-2)1

and 1 , as (y-2) is the same number as (y-2)1

The product is therefore, (y-2)(1+1) = (y-2)2

Equation at the end of step

3

:

4 • (y - 2)2 = 0

STEP

4

:

Equations which are never true:

4.1 Solve : 4 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

4.2 Solve : (y-2)2 = 0

(y-2) 2 represents, in effect, a product of 2 terms which is equal to zero

For the product to be zero, at least one of these #

One solution was found :

y = 2