If (2y - 3)^2 - 2(2y - 3) - 24 = 0 and y > 0, what is the value of y? — SAT

If $(2y - 3)^2 - 2(2y - 3) - 24 = 0$ and $y > 0$ , what is the value of $y$ ?

\frac{9}{2}

Substituting

u = 2y - 3

transforms the original equation into the standard quadratic form

u^2 - 2u - 24 = 0

. Factoring this equation yields

(u - 6)(u + 4) = 0

, giving the solutions

u = 6

and

u = -4

. Substituting

2y - 3

back for

u

results in two equations:

2y - 3 = 6

(which simplifies to

y = \frac{9}{2}

) and

2y - 3 = -4

(which simplifies to

y = -\frac{1}{2}

). Since the problem specifies that

y > 0

, the negative value must be discarded, leaving the correct value as

\frac{9}{2}

Use substitution to simplify the equation by letting

u = 2y - 3

u^2 - 2u - 24 = 0

This substitution reduces the equation to a standard quadratic form, making it easier to factor.

Factor the quadratic equation

u^2 - 2u - 24 = 0

(u - 6)(u + 4) = 0

, which gives

u = 6

u = -4

Factoring helps identify the potential values for the substituted expression

u

Substitute

2y - 3

back for

u

and solve both resulting linear equations for

y

2y - 3 = 6 \implies 2y = 9 \implies y = \frac{9}{2}

and

2y - 3 = -4 \implies 2y = -1 \implies y = -\frac{1}{2}

This step converts the solutions for the intermediate variable

u

back into solutions for the original variable

y

Apply the given constraint

y > 0

to identify the final solution.

y = \frac{9}{2}

because

-\frac{1}{2}

is not greater than

0

The question specifies that

y

must be strictly positive, so we must discard any negative values.

Solving quadratic equations using substitution and factoring under constraints

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