x^2 - 8x + 52 को पूर्ण वर्ग बनाकर हल करें

Problem

Solve by completing the square:

$x^2 - 8x + 52 = 0$

Step 1: Isolate the $x$ Terms

To complete the square, move the constant term away from the $x^2$ and $x$ terms. Subtract $52$ from both sides:

$x^2 - 8x = -52$

Step 2: Create a Perfect Square

Take half of the coefficient of $x$ and square it. Half of $-8$ is $-4$, and

$(-4)^2 = 16$

Add $16$ to both sides:

$x^2 - 8x + 16 = -52 + 16$

$x^2 - 8x + 16 = -36$

Step 3: Rewrite the Trinomial

The left side is now a perfect square trinomial:

$x^2 - 8x + 16 = (x - 4)^2$

So the equation becomes

$(x - 4)^2 = -36$

Step 4: Take Square Roots

Take the square root of both sides, remembering both the positive and negative possibilities:

$x - 4 = \pm \sqrt{-36}$

Since

$\sqrt{-36} = 6i$

we get

$x - 4 = \pm 6i$

Step 5: Solve for $x$

Add $4$ to both sides:

$x = 4 \pm 6i$

So the solutions are

$x = 4 - 6i \quad \text{and} \quad x = 4 + 6i$