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If the range of quadratic polynomial $f(x)= 2x^{2}+3x+\frac{a}{b}$ is $[4, \infty)$ and $a$ and $b$ are co prime. Find $a+b$ .

Options are $50,46,49,58$.

What I did was I made a rough graph and the roots are complex. So I saw Discriminant is less than zero. Then another method I tried was I made some cases like if $a$ and $b$ are co prime then either both are odd which lead to both will be prime or one is even and one is odd. Any method I try I get stuck. Please help me . Thank you for your time .

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$$f(x)=2\left(x^2+\frac{3}{2}x\right)+\frac{a}{b}=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}+\frac{a}{b}$$

The minimum value is

$$\frac{-9}{8}+\frac{a}{b}=4$$

$$\frac{a}{b}=\frac{41}{8}$$

$a=41$ and $b=8$.

$a+b=49$.

Another method:

The minimum value of $f(x)$ is $4$. So the graph has the vertex with $y$-coordinate equals to $4$ and hence the equation $f(x)-4=0$ has a double root.

$f(x)-4=2x^2+3x+\frac{a}{b}-4$.

$$3^2-4(2)\left(\frac{a}{b}-4\right)=0$$

$$\frac{a}{b}=\frac{41}{8}$$

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  • $\begingroup$ Thank you for your time . Just wanted to know that can you tell how to proceed with my methods $\endgroup$ – Abhishek Mehta May 25 '17 at 8:00

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