CHAPTER 7. SOLVINGQUADRATIC EQUATIONS 7.4
SOLUTION
Step 1 : Determine whether theequation can be factorised
The expression cannotbe factorised. Therefore, the general quadraticformula
must be used.
Step 2 : Identify the coefficients in the equation for usein the formula
From the equation:
a = 1
b =− 5
c = 8
Step 3 : Apply the quadratic formula
x =
−b±
√
b^2 − 4 ac
2 a
(7.14)
=
−(−5)±
�
(−5)^2 − 4(1)(8)
2(1)
(7.15)
=
5 ±
√
− 7
2
(7.16)
(7.17)
Step 4 : Write the final answer
Since the expression under the square root is negative these are not real solutions
(
√
− 7 is not a real number). Therefore there are no real solutions to the quadratic
equation x^2 − 5 x + 8 = 0. This means that the graph of the quadratic function
f(x) = x^2 − 5 x + 8 has no x-intercepts, but that theentire graph lies above the
x-axis.
See video: VMezc at http://www.everythingmaths.co.za
Exercise 7 - 3
Solve for t using the quadratic formula.
- 3 t^2 + t− 4 = 0
- t^2 − 5 t + 9 = 0
- 2 t^2 + 6t + 5 = 0
- 4 t^2 + 2t + 2 = 0
5.− 3 t^2 + 5t− 8 = 0
6.− 5 t^2 + 3t− 3 = 0
- t^2 − 4 t + 2 = 0