What is the value of x? (1) 3x^2 + 8x + 16/3 = 0 (2) 15x^2 + 80/3 = -4

1 can be rewritten as (3x+4)^2 so we can find X.
2. can be rewritten as:
45x^2 + 80 +120x = 0
divide by 5:
9x^2 + 16+24x=0
basically the same equation
we can factor out (3x+4)^2 = 0

D

for those like me who did not find square immediately:

for equation in form:
\(ax^2+bx+c=0\)

has the unique solution if:
\(b^2-4ac=0\)

Be careful with the \(+\) and \(-\) signs though. I often have confused myself with signs.

(1) 3x^2 + 8x + 16/3 = 0 can be rewritten as
9x^2 + 24x + 16 = 0.. OR (3x+4)^2 = 0
This gives us x=-4/3. Sufficient.

(2) 15x^2 + 80/3 = -40x can be rewritten as
15x^2 + 40x + 80/3 = 0 .. Dividing this by 5, we get
5x^2 + 8x + 16/3 = 0. This is same as the equation in statement 1, and will give us same value of x. Sufficient.

Hence D answer

Target question: What is the value of x?

Statement 1: 3x² + 8x + 16/3 = 0
Let's first eliminate the fractions by multiplying both sides of the equation by 3
When we do this, we get: 9x² + 24x + 16 = 0
Factor to get: (3x + 4)(3x + 4) = 0
So, it must be the case that 3x + 4 = 0
So, 3x = -4
x = -4/3
So, the answer to the target question is x = -4/3
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 15x² + 80/3 = -40x
Let's first eliminate the fractions by multiplying both sides of the equation by 3
When we do this, we get: 45x² + 80 = -120x
Add 120x to both sides: 45x² + 120x + 80 =0
Divide both sides by 5 to get: When we do this, we get: 9x² + 24x + 16 = 0
This is the same equation we had in statement 1
So, the answer to the target question is x = -4/3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

(1) \(3x^2 + 8x + \frac{16}{3} = 0\)
9x^2 +24x+ 16 =0

upon solving we get x=-4/3
Clearly sufficient

(2) \(15x^2 + \frac{80}{3} = -40x\)

9x^2 +24X +16=0
same eqn same x value

THerefore IMO D

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