how to solve the following equations giving your answer correct to 2 decimal places where necessary
(a) (x+1)2=9
Since a constant multiplier is not usually written behind a binomial, I suspect you meant:
(x+1)^2 = 9
then x+1 = ±√9 = ±3
x+1 = 3 ----> x = 2
or
x+1 = -3 ---> x = -4
(x+1)2=9
2x + 2 = 9
2x = 7
x = 3.5
To solve the equation (x+1)^2 = 9, you can follow these steps:
Step 1: Expand the equation
(x+1)^2 = 9
(x+1)(x+1) = 9
x^2 + 2x + 1 = 9
Step 2: Rearrange the equation
x^2 + 2x + 1 - 9 = 0
x^2 + 2x - 8 = 0
Step 3: Solve the quadratic equation. To do this, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 2, and c = -8. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(-8))) / (2(1))
Step 4: Simplify the equation
x = (-2 ± √(4 + 32)) / 2
x = (-2 ± √36) / 2
x = (-2 ± 6) / 2
Step 5: Calculate the roots
x1 = (-2 + 6) / 2
x1 = 4 / 2
x1 = 2
x2 = (-2 - 6) / 2
x2 = -8 / 2
x2 = -4
So, the solutions to the equation (x+1)^2 = 9 are x = 2 and x = -4.