The square of 3 less than a number x is equal to 16. What is the greater of the two possib… — ACT

The square of $3$ less than a number $x$ is equal to $16$ . What is the greater of the two possible values for $x$ ?

The word problem translates to the equation

(x-3)^2 = 16

. Expanding the left side and subtracting 16 from both sides gives the standard quadratic equation

x^2 - 6x - 7 = 0

. This factors into

(x-7)(x+1) = 0

, yielding two solutions:

x = 7

and

x = -1

. The greater of these two values is 7.

Translate the verbal description into an algebraic equation.

(x-3)^2 = 16

'3 less than a number

x

' is written as

x - 3

, and its square is set equal to 16.

Expand the squared binomial and rearrange the terms into standard quadratic form

ax^2 + bx + c = 0

x^2 - 6x - 7 = 0

Expanding

(x-3)^2

yields

x^2 - 6x + 9

. Subtracting 16 from both sides sets the equation to zero.

Factor the quadratic equation over the integers.

(x - 7)(x + 1) = 0

We find two integers that multiply to

-7

and add to

-6

, which are

-7

and

1

Solve for

x

by setting each factor to zero, then select the greater value.

x = 7

(since the solutions are

7

and

-1

)

Setting the factors to zero gives

x - 7 = 0 \implies x = 7

and

x + 1 = 0 \implies x = -1

. The larger of these two values is

7

Solving Quadratic Equations by Factoring

Estimated Time:1m 0s

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