Question

If x²+ax-4=0 is a perfect square,find the value of a

Answer 1

To determine the value of \( a \) such that the quadratic equation \( x^2 + ax - 4 = 0 \) is a perfect square, we can express it in the form \( (x + b)^2 \) for some \( b \).

Expanding \( (x + b)^2 \) gives us:

\[
(x + b)^2 = x^2 + 2bx + b^2
\]

We want this to match the form \( x^2 + ax - 4 \). Thus, we can compare coefficients:

- The coefficient of \( x \) gives us:
\[
2b = a
\]
- The constant term gives us:
\[
b^2 = -4
\]

The equation \( b^2 = -4 \) leads us to conclude that there are no real values of \( b \) because the square of a real number cannot be negative. However, if we are allowing complex values for \( b \), we can solve for \( b \):

\[
b = \pm 2i
\]

Substituting \( b = 2i \) into \( 2b = a \):

\[
a = 2(2i) = 4i
\]

Substituting \( b = -2i \):

\[
a = 2(-2i) = -4i
\]

Thus, the values of \( a \) that make the original quadratic a perfect square (in the complex number domain) are:

\[
a = 4i \quad \text{or} \quad a = -4i
\]

In conclusion, the values of \( a \) are \( \boxed{4i} \) and \( \boxed{-4i} \) if complex values for \( a \) are permissible.