gmattesttaker2 wrote:
How many times does graph y = ax^2 + bx + c intersect the x-axis?
1) a > 0
2) c < 0
OA: C
For any equation of the form y = ax² + bx + c, the DISCRIMINANT is equal to b² - 4ac.
If the discriminant is POSITIVE, then the graph of the equation intersects the x-axis TWICE.
If the discriminant is ZERO, then the graph of the equation intersects the x-axis ONCE.
If the discriminant is NEGATIVE, then the graph of the equation DOES NOT INTERSECT the x-axis.
Question stem, rephrased: Is b² - 4ac positive, zero, or negative?
Statement 1: a > 0
Test one case that satisfies BOTH statements.
Case 1: If b=1, a=1, and c=-1, then b² - 4ac = 1² - (4*1)(-1) = 5, which is positive.
Test one case that satisfies ONLY STATEMENT 1.
Case 2: If b=1, a=1, and c=1, then b² - 4ac = 1² - 4*1*1 = -3, which is negative.
INSUFFICIENT.
Statement 2: c < 0
Case 1 also satisfies statement 2.
In Case 1, the discriminant is positive.
Test one case that satisfies ONLY STATEMENT 2.
Case 3: If b=1, a=-1, and c=-1, then b² - 4ac = 1² - (4)(-1)(-1) = -3, which is negative.
INSUFFICIENT.
Statements combined:
Since the square of a value cannot be negative, b² ≥ 0.
Since a>0 and c<0, 4ac < 0.
Thus:
b² - 4ac = nonnegative - negative = nonnegative + positive = positive.
SUFFICIENT.
The correct answer is
C.
I wouldn't worry too much about this problem.
Questions about parabolas appear on the GMAT only rarely.
That said, my solution above combines two concepts discussed briefly in the OG13:
In its discussion of quadratics on page 125, the OG13 mentions that the number of SOLUTIONS depends on the sign of b² - 4ac.
In its discussion of parabolas on page 139, the OG13 mentions that the x-interecepts occur when the quadratic is EQUAL TO 0.
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