Question

Consider the equation: 12x=13-x^212x=13−x 2 12, x, equals, 13, minus, x, squared 1) Rewrite the equation by completing the square. Your equation should look like (x+c)^2=d(x+c) 2 =dleft parenthesis, x, plus, c, right parenthesis, squared, equals, d or (x-c)^2=d(x−c) 2 =dleft parenthesis, x, minus, c, right parenthesis, squared, equals, d. 2) What are the solutions to the equation

Answer 1

Answer:

$(x + 6)^2 = 49$

$x = 1$    or $x = -13$

Step-by-step explanation:

Given

$12x = 13 - x^2$

Using Completing the Square

$12x = 13 - x^2$ ---- Add $x^2$ to both sides

$x^2 + 12x = 13 - x^2 + x^2$

$x^2 + 12x = 13$

Divide the coefficient of x by 2; then add the square to both sides

$x^2 + 12x + 6^2 = 13 + 6^2$

$x^2 + 12x + 36 = 13 + 36$

$x^2 + 12x + 36 = 49$

Factorize

$x^2 + 6x + 6x + 36 = 49$

$x(x + 6) + 6(x + 6) = 49$

$(x + 6)(x + 6) = 49$

$(x + 6)^2 = 49$

Hence, the equation is $(x + 6)^2 = 49$

Solving further

Take square root of both sides

$(x + 6) = \sqrt{49}$

$x + 6 = \±7$

$x = \±7- 6$

This implies that

$x = 7 - 6$     or $x = -7 -6$

$x = 1$    or $x = -13$

HEnce, the solutions are $x = 1$    or $x = -13$