\(\displaystyle x^{2} +Bx = C\) can be rewritten as \(\displaystyle x^{2} +Bx - C = 0\)

If Statement 1 holds, then the equation can be rewritten as \(\displaystyle (x+m) (x+n)=0\). This equation has solution set \(\displaystyle \left \{ -m,-n\right \}\), which comprises two real numbers.

If Statement 2 holds, the discriminant \(\displaystyle B^{2} -4 \cdot1\cdot(-C)= B^{2} +4C\) is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

The number of real solutions of the equation \(\displaystyle Ax^{2} + Bx + C = 0\) depends on whether discriminant \(\displaystyle B^{2} - 4A C\) is positive, zero, or negative; since \(\displaystyle A = 1\), this becomes \(\displaystyle B^{2} - 4C\).

If we only know that \(\displaystyle C\) is a perfect square, then we still need to know \(\displaystyle B\) to find the number of real solutions. For example, let \(\displaystyle C = 1\), a perfect square. Then the discriminant is \(\displaystyle B^{2} - 4\), which can be positive, zero, or negative depending on \(\displaystyle B\).

But if we know  \(\displaystyle C = \left ( \frac{B}{2} \right ) ^{2}\) , then the discriminant is 

\(\displaystyle B^{2} - 4C = B^{2} - 4 \cdot \left ( \frac{B}{2} \right ) ^{2} = B^{2} - 4 \cdot \frac{B^{2}}{4} = B^{2} - B^{2} = 0\)

Therefore, \(\displaystyle x^{2} + Bx + C = 0\) has one real solution.

The sign of the discriminant of the quadratic expression answers this question; here, the discriminant is

\(\displaystyle (-3A)^{2} - 4AB\),

or

\(\displaystyle 9A^{2} - 4AB\)

If we assume Statement 1 alone, this expression becomes

\(\displaystyle 9A^{2} - 4AB = 9A^{2} - 4A (2A) = 9A^{2} - 8A^{2} = A^{2}\)

Since we can assume \(\displaystyle A\) is nonzero, \(\displaystyle A^{2} > 0\). This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

\(\displaystyle 9A^{2} - 4AB = 9A^{2} - 4A \left ( A + 2\right ) = 9A^{2} - 4A^{2}-8A = 5A^{2}-8A\)

The sign of \(\displaystyle 5A^{2}-8A\) can vary.

Case 1: \(\displaystyle A = 1\)

Then \(\displaystyle 5A^{2}-8A = 5 \cdot 1^{2}-8\cdot 1 = 5 - 8 = -3 < 0\)

giving the equation two imaginary solutions.

Case 2: \(\displaystyle A = 2\)

Then \(\displaystyle 5A^{2}-8A = 5 \cdot 2^{2}-8\cdot 2 = 20 - 16 = 4 > 0\)

giving the equation two real solutions.

Therefore, Statement 1, but not Statement 2, is enough to answer the question.

The quadratic expression can be factored as \(\displaystyle (x+? )(x+? )\), replacing the question marks with integers whose product is 8 and whose sum is \(\displaystyle -6\). These integers are \(\displaystyle -2,-4\), so the equation becomes:

\(\displaystyle x^{2} - 6x + 8 = 0\)

\(\displaystyle (x-2 )(x-4 ) = 0\)

Set each linear binomial to 0 and solve:

\(\displaystyle x - 2 = 0 \Rightarrow x = 2\)

\(\displaystyle x - 4 = 0 \Rightarrow x = 4\)

Therefore, for the statement to be true, either \(\displaystyle x = 2\) or \(\displaystyle x = 4\). Each of Statement 1 and Statement 2, taken alone, leaves other possible values of \(\displaystyle x\). Taken together, however, they are enough, since the only two positive even integers less than 6 are 2 and 4.

The two statements together are insufficient. If both are assumed, then \(\displaystyle x\) can be 2, 4, or 6. 

If \(\displaystyle x = 4\) the statement is true:

\(\displaystyle x^{2} - 10x +24 = 0\)

\(\displaystyle 4^{2} - 10 \cdot 4 +24 = 0\)

\(\displaystyle 16 - 40 +24 = 0\)

\(\displaystyle 0 = 0\)

But if \(\displaystyle x = 2\) the statement is false:

\(\displaystyle x^{2} - 10x +24 = 0\)

\(\displaystyle 2^{2} - 10 \cdot 2 +24 = 0\)

\(\displaystyle 4 - 20 +24 = 0\)

\(\displaystyle 8 = 0\) 

This is a quadratic formula problem. Use equation \(\displaystyle \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). For our problem, \(\displaystyle a=5, b=-4, c=3.\)  Plug these values into the equation, and simplify:

\(\displaystyle \frac{2 \pm \sqrt{(-4)^{2} - 4(5)(-3)}}{2(5)}\)     \(\displaystyle =\frac{2 \pm \sqrt{16 +60}}{10}=\frac{2 \pm \sqrt{76}}{10}=\frac{1 \pm \sqrt{19}}{5}\). Here, we simplified the radical by \(\displaystyle \sqrt{76}=\sqrt{4*19}=\sqrt{4}\sqrt{19}=2\sqrt{19}.\)