Question 200·200 Super-Hard SAT Math Questions·Advanced Math

0 / 200 Questions completed

Let $a$ and $b$ be integers. The quadratic equation $x^2+ax+b=0$ has no real solutions.

In addition, $x+2$ is a factor of the polynomial $x^3+ax^2+bx+16$ .

Which choice is the greatest possible value of $b$ ?

A stated factor gives an input that makes a polynomial zero, linking its unknown coefficients. A quadratic with no real solutions has a negative discriminant. Use the factor to express one coefficient in terms of the other, then compare the two sides of the discriminant inequality in Desmos. The boundary does not count because the inequality is strict.

Hints

Hint 1
A factor makes the whole polynomial zero wherever that factor is zero. Set $x+2$ equal to zero, then use that input in the cubic to find a relationship between $a$ and $b$ .
Hint 2
The discriminant is the number under the square root in the quadratic formula. It must be negative for the quadratic to have no real solutions. How can you use the factor relationship to write that condition with only $a$ ?
Hint 3
Graph both sides of the inequality, using $x$ to stand for $a$ . Where is the parabola below the line? Remember that $a$ must be an integer when you choose its greatest allowed value.

Step-by-step

Approach 1: Link the coefficients, then graph the cutoff
Step 1
Use the given factor
The factor $x+2$ equals zero at $x=-2$ , so the cubic must equal zero there:
$(-2)^3+a(-2)^2+b(-2)+16=0.$
Simplify:
$8+4a-2b=0.$
Step 2
Express b using a
Move $2b$ to the other side:
$8+4a=2b.$
Divide by $2$ :
$b=2a+4.$
So larger $a$ means larger $b$ ; you need the greatest $a$ that the quadratic allows.
Step 3
Apply the no-real-solutions condition
The discriminant of $x^2+ax+b$ is $a^2-4b$ . If it's negative, its square root isn't real. So no real solutions requires the discriminant to be strictly below zero:
$a^2-4b<0.$
Step 4
Use the factor relationship
Replace $b$ with $2a+4$ in the inequality:
$a^2-4(2a+4)<0.$
Step 5
Put the inequality in graph-ready form
Add $4(2a+4)$ to both sides:
$a^2<4(2a+4).$
Now you can compare the two sides as graphs.
Step 6
Find the greatest allowed integer
Type $y=x^2$ and $y=4(2x+4)$ , with $x$ standing for $a$ . Desmos shows crossings near $x=-1.657$ and $x=9.657$ ; click the right one to see about $(9.657,93.255)$ . The parabola is below the line between the crossings, so the greatest integer $a$ allowed is $9$ , not $10$ .
Step 7
Find the greatest b
Type $2(9)+4$ . Desmos shows $22$ , the greatest possible value of $b$ . Since $a=9$ lies between the crossings, the quadratic has no real solutions. Choice C.

Approach 2: Find the integer cutoff with exact squares
Step 1
Expand the boundary expression
You can find the cutoff without estimating a graph crossing. Start from $a^2<4(2a+4)$ and distribute:
$a^2<8a+16.$
Step 2
Move the a term
Subtract $8a$ from both sides:
$a^2-8a<16.$
This prepares the left side for a perfect square.
Step 3
Complete the square
Add $16$ to both sides:
$a^2-8a+16<32.$
Step 4
Write the square
Because $a^2-8a+16$ is a perfect square, rewrite it:
$(a-4)^2<32.$
Step 5
Use the integer condition
Since $5^2=25<32<36=6^2$ , the greatest integer $a-4$ can be is $5$ . That gives $a=9$ . A value of $6$ for $a-4$ would make the square too large.
Step 6
Return to the requested coefficient
Use the factor relationship $b=2a+4$ . Type $2(9)+4$ ; Desmos shows $22$ for the greatest possible $b$ . Choice C.

Question 200·200 Super-Hard SAT Math Questions·Advanced Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this