When using the quadratic formula to solve a quadratic equation.........?

Positive discriminant

(x-a)(x-b) is and eqn where both a and b can be any real nos. eg (x-3)(x-4) simplify = x^2 -7x +12

(x-2/3)(x-7/5) simplify

(in fact general form (ax+b)(cx +d) where a,b,c,d are real.)

2. (ax+b)^2 =0 has equal roots i.e d=0

(2x +3)^2 = 0 solve to get eqn 4x^2 +12x +9 =0

3. if you know complex then (x +a)(x+b)= 0 has d -ive if a,b are complex or imaginary

otherwise x^ 2= -ive no will also have d <0

eg X^2 +9 =0 or (x+3i)(x-3i)=0

on graph the parabola cuts x axis at two point d>0

one point the turning point d= 0

no point d <0

To make the discriminant what you want just pick out your (b) well.

For instance... a=1; b=10; c=1... the discriminant is: 96 (positive)

For instance... a=1; b=1; c=1... the discriminant is: -3 (negative)

For instance... a=1; b=2; c=1... the discriminant is: 0 (zero)

When the discriminant is:

positive... the graph has (2) real zeros.

negative... the graph has (2) imaginary zeros.

zero... the graph has (1) real zero.

English...

Zeros are where the graph crosses the x-axis. If a graph crosses the x-axis twice, it has two zeros. If the graph touches the x-axis once (x^2 for example), it has one zero. If the graph has imaginary zeros, then the graph doesn't cross at all. Meaning the graph is always above the x-axis.

nicely the formula for fixing an quadratic equation is -b + - the sq. root of b squared + 4 x A rewrite the equation like this 8x + x -5 8x=A x=b and -5 is c -x + and minus x to the 2d skill + 4 x 8 x -5 be squared equals 8x8 that's sixty 4 + 4 x a x -5 an excellent 20 x X 20 x a million is 20 the sq. root of 20 is 4.40 seven + -8 = -3.fifty 3 then minus is is 11.fifty 3 so that you've 2 solutions 11.fifty 3 effective and -3.fifty 3 unfavorable volume thats how i do quadratic equations I dont understand if its top you would opt to envision with someone who's familiar with more desirable about the stuff like a math instructor

if b^2 - 4ac = 0, it means that there is exactly one solution.

if you graph y = ax^2 + bx + c, it is a parabola. the vertex of that parabola will touch exactly one point (the solution) on the x-axis.

if b^2 - 4ac = a positive number, it means that there are two solutions. if you graph the parabola, the vertex will be below the x-axis, which means that the parabola will cross two points (2 solutions) on the x-axis

if b^2 - 4ac = negative, the vertex of the parabola will be above the x-axis. this means that no points touch the x-axis, so therefore there is no real solution. also, for this case, if b^2 - 4ac is negative, the quadratic equation takes the square root of if b^2 - 4ac. you can not take the square root of a negative number, that is also why there is no real solutions, just imaginary

well for the one where the discriminant is zero, couldnt you just make a, b, and c all equal zero? or is that like cheating? for the negative one, you could do like (3)^2 -4(4)(5), so that would be 4x^2 + 3x + 5 = 0. For the positive one you could do (6)^2 -4(4)(2), so that would be 4x^2 + 6x + 2 = 0. Hope thats what you were looking for!

The quadratic formula is ... : X equals the opposite of B plus/minus the square root of B squared minus 4AC, all over 2A. A = The whole number; B= The variable amount; C= The variable squared amount.

It's really hard to explain, because I don't know how to type the symbols on my computer... sorry...

wow....sucks for you, im done with all this algebra crap for the summer. good luck with this one