5x^2+8x+7=0
(7)^1/2y^2-6y-13(7)1/2=0
2x^2+x-1=0
4/3x^2-2x+3/4=0
2007-07-20 05:05:29 · 4 answers · asked by littlesoutherner12 1 in Science & Mathematics ➔ Mathematics
All of these eqations have solutions. Some are called real solutions and others are not real or imaginary. If you want to know which is real or a whole number then use factoring but if you want to get the solution real or imaginary then use the quadratic equation
x= -b+-(sqrt(b^2-4ac))/2a
where
a= x^2 coeff
b= x coeff
c=constant
2007-07-20 05:15:36 · answer #1 · answered by 037 G 6 · 1⤊ 0⤋
Use the dicriminant (b^2 - 4ac) to determine what sort of solutions a quadratic equation has.
(The letters a,b,and c come from: ax^2 + bx + c = 0. (substitute the numbers for a, b, and c)
If the discriminant is:
positive, 2 real solutions
0, 1 real solution
negative, no real solutions (the solutions are imaginary)
2007-07-20 12:27:23 · answer #2 · answered by pgd_malaka 6 · 1⤊ 0⤋
If it is an equation there will be solution. Some can be easily solved using the "splitting the middle term" method. Others can be solved using the formula for general QUADRATIC EQUATION. This equation is: ax^2 + bx + c
...........Here x can be solved using the formula:
.......................x = - b ± Ѵ(b² - 4ac) / 2a where a is the co efficient of x², b,the coefficient of x, and c is the constant. These roots (solutions) may be either real or imaginary. The term (b² - 4ac) will decide whether it is a real one or imaginary one. It is called descriminant.
2007-07-20 12:35:46 · answer #3 · answered by Joymash 6 · 0⤊ 0⤋
All of them have solutions -- see the quadratic equation.
2007-07-20 12:23:02 · answer #4 · answered by miggitymaggz 5 · 0⤊ 0⤋