Respuesta :
Sure, to solve the quadratic equation \( x^2 - 12x + 11 = 0 \) using the completing the square method, we need to rewrite the equation in the form \( (x - h)^2 = k \).
First, we move the constant term to the right side:
\( x^2 - 12x = -11 \)
Next, we need to complete the square. To do this, we halve the coefficient of \( x \) (which is -12), square it, and add it to both sides:
\( x^2 - 12x + (-6)^2 = -11 + (-6)^2 \)
\( x^2 - 12x + 36 = -11 + 36 \)
\( x^2 - 12x + 36 = 25 \)
Now, we can rewrite the left side as a perfect square:
\( (x - 6)^2 = 25 \)
Taking the square root of both sides:
\( x - 6 = \pm \sqrt{25} \)
\( x - 6 = \pm 5 \)
Adding 6 to both sides:
\( x = 6 \pm 5 \)
So, the solutions are \( x = 6 + 5 = 11 \) and \( x = 6 - 5 = 1 \).
First, we move the constant term to the right side:
\( x^2 - 12x = -11 \)
Next, we need to complete the square. To do this, we halve the coefficient of \( x \) (which is -12), square it, and add it to both sides:
\( x^2 - 12x + (-6)^2 = -11 + (-6)^2 \)
\( x^2 - 12x + 36 = -11 + 36 \)
\( x^2 - 12x + 36 = 25 \)
Now, we can rewrite the left side as a perfect square:
\( (x - 6)^2 = 25 \)
Taking the square root of both sides:
\( x - 6 = \pm \sqrt{25} \)
\( x - 6 = \pm 5 \)
Adding 6 to both sides:
\( x = 6 \pm 5 \)
So, the solutions are \( x = 6 + 5 = 11 \) and \( x = 6 - 5 = 1 \).
Answer:
x^2 - 12x = -11
or, x^2 - 2×x×6 +6^2 = -11 + 6^2
or, (x - 6)^2 = 25
or, x - 6 = ± 5
taking + ve,
x = 11
taking - ve,
x = 1
