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.
Solve,correct to 2 decimal places the equation 4x^2 = 11x + 23
1 answer