Question

Number of integer values of k for which the quadratic equation $2x^2+kx-4=0$ will have two rational solutions is

• A. $1$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Dividing the equation by the coefficient of $x^2$ i.e. 2 we get
$x^2+\frac{kx}{2}-2=0$
${(x+\frac{k}{4})}^2-\frac{k^2}{16}-2=0$
${(x+\frac{k}{4})}^2=\frac{k^2+32}{16}$
Hence for rational roots, $\frac{k^2+32}{16}$ has to be a perfect square.
We get a perfect square at $k=\pm 2$ for ($\frac{k^2+32}{16}$) i.e. $\frac{36}{16}$ which becomes $\frac{6}{4}$ upon removing the square

We get a perfect square at $k=\pm 7$ for ($\frac{k^2+32}{16}$) i.e. $\frac{81}{16}$ which becomes $\frac{9}{4}$ upon removing the square

Hence the number of integral values of k is $4(2,-2,7,-7)$