Let integer $$\alpha$$ be one of the roots of $$ax^2 + 2x + 3 = 0$$ and integer $$\beta$$ be one of the roots of $$5x^2 + bx + 8 = 0$$. It is also given that $$\alpha = \beta^2$$. Which of the following statement is true if $$b = -14$$?

$$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ 4^2+32\times\ -\frac{11}{16}\times\ 4$$It is given that b=-14.

Then the eq $$5x^2 + bx + 8 = 0$$ becomes $$5x^2 -14x + 8 = 0$$.

$$x=\frac{-b\pm\ \sqrt{\ b^2-4ac}}{2a}=\frac{14\pm\ \sqrt{\ 196-160}}{10}$$

$$=\frac{14\pm\ 6}{10}$$

$$x=2\ or\ 0.8$$

Hence $$\beta=2$$.

If $$\beta=2$$, Then $$\alpha = \beta^2$$ = $$4$$.

As $$\alpha$$ is a root of $$ax^2 + 2x + 3 = 0$$

$$4^2a+8+3=0$$

$$a=-\frac{11}{16}$$.

By verifying options, Option D satisfies for the a & b values.

= $$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ \left(-14\right)^2+32\times\ -\frac{11}{16}\times\ -14$$

$$=\ -275$$

Hence Option (D) is the answer.