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Algebra Difficulty 2.8 Junior Find the answer

Let F=6x2+16x+3m6F=\frac{6x^2+16x+3m}{6} be the square of an expression which is linear in xx. Then mm has a particular value between:
(A) 3 and 4\text{(A) } 3 \text{ and } 4(B) 4 and 5\text{(B) } 4 \text{ and } 5(C) 5 and 6\text{(C) } 5 \text{ and } 6(D) −4 and −3\text{(D) } -4 \text{ and } -3(E) −6 and −5\text{(E) } -6 \text{ and } -5

Multiple choice: answer with the letter of the option you want.

Solution

The quadratic can also be written as
x2+83x+m2x^2 + \frac{8}{3}x + \frac{m}{2}
In order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4. Thus,
m2=(83)2⋅14\frac{m}{2} = (\frac{8}{3})^2 \cdot \frac{1}{4}
m2=169\frac{m}{2} = \frac{16}{9}
m=329m = \frac{32}{9}
The answer is (A)\boxed{\textbf{(A)}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.