Solve,correct to 2 decimal places the equation 4x^2 = 11x + 23
To solve the equation 4x^2 = 11x + 23, we can rearrange it to the form 4x^2 - 11x - 23 = 0 and then solve for x using the quadratic formula.
The quadratic formula is given by x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.
For our equation, a = 4, b = -11, and c = -23. Substituting these values into the quadratic formula, we have:
x = (-(-11) ± sqrt((-11)^2 - 4(4)(-23))) / (2(4))
x = (11 ± sqrt(121 + 368)) / 8
x = (11 ± sqrt(489)) / 8
Therefore, the solutions to the equation are x = (11 + sqrt(489)) / 8 and x = (11 - sqrt(489)) / 8. These values can be approximated to 2 decimal places.
Calculating these values, we get:
x ≈ (11 + sqrt(489)) / 8 ≈ 4.06
x ≈ (11 - sqrt(489)) / 8 ≈ -1.56
So the solutions to the equation 4x^2 = 11x + 23, correct to 2 decimal places, are x ≈ 4.06 and x ≈ -1.56.
To solve the equation 4x^2 = 11x + 23, follow these steps:
Step 1: Set the equation equal to zero by subtracting 11x and 23 from both sides:
4x^2 - 11x - 23 = 0
Step 2: Use the quadratic formula to find the solutions for x. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4, b = -11, and c = -23. Substituting these values into the quadratic formula, we get:
x = (-(-11) ± √((-11)^2 - 4 * 4 * -23)) / (2 * 4)
x = (11 ± √(121 + 368)) / 8
x = (11 ± √489) / 8
Step 3: Simplify the square root:
x = (11 ± √489) / 8
x ≈ (11 + √489) / 8 ≈ 4.45
x ≈ (11 - √489) / 8 ≈ -0.95
Therefore, the solutions to the equation 4x^2 = 11x + 23, correct to 2 decimal places, are approximately x ≈ 4.45 and x ≈ -0.95.