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How do you know if a quadratic equation will have one, two, or no solutions?

How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide  me with one or two solutions with which they must create a quadratic equation.

USE THE DISCRIMINANT TO FIND OUT THE NATURE OF THE ROOTS, NOW THE NATURE OF THE ROOTS WILL TELL YOU IF THE ANSWER IS RATIONAL, IRRATIONAL, IMAGINARY, OR IF THERE IS ONE OR TWO ANSWER.
SINCE YOU ARE ONLY WORRIED ABOUT THE NUMBER OF SOLUTIONS THEN THAT'S ALL WE'LL FOCUS ON.

THE DISCRIMINANT IS B^2 -4AC

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REGARDING How do you find a quadratic equation if you are only given the solution?

YOU NEED SUM OF THE ROOTS AND PRODUCT OF THE ROOTS

SUM OF THE ROOTS = -B/A
PRODUCT OF THE ROOTS = C/A

SO LETS SAY YOU HAVE A QUADRATIC W SOLUTIONS (SAME THING AS ROOTS)  X=5 AND X= 7

PRODUCT OF THE ROOTS WOULD BE 35 (MULTIPLY THE ANSWERS)

SO.... SUM OF THE ROOTS IS  -B/A, BUT WE KNOW ITS 12..  WE CAN SAY -B/A = 12/1.   I PUT THE 12 OVER 1 BC ITS A WHOLE NUMBER AND I JUST LIKE TO COMPARE FRACTIONS TO OTHER FRACTIONS, ITS A LITTLE EASIER.

THE PRODUCT OF THE ROOTS IS C/A OR 35, SO C/A = 35/1.

C/A = 35/1   -B/A = 12/1

WE CAN THEN SAY THAT A = 1, B = -12 (WATCH YOUR SIGNS) AND C = 35
THE QUADRATIC THAT GOES W THIS WOULD BE
Y=X^2 -12X+35.

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Is it possible to have different quadratic equations with the same solution?

YES, A SOLUTION IS JUST WHERE THE QUADRATIC CROSSES OVER THE X AXIS.

SO EASY EXAMPLE WOULD BE Y=X^2 AND Y=5X^2.  THEY ARE DIFFERENT QUADRATICS AND THEY BOTH HAVE ONLY ONE SOLUTION, ZERO

Discriminant of a quadratic equation provides you with the indication of the number of solutions.

For the equation ax2+bx+c=0, D=b2-4ac is the discriminant.

1) D>0; there are two distinct solutions.
x1={-b-√(b2-4ac)}/(2a)=(-b-√D)/(2a)
x2=(-b+√D)/(2a)

Example:

x2-5x+6=0; D=52-4*6*1=25-24=1;
x1=(5-√1)/2=2; x2=(5+√1)/2=3;

2) D=0; There is one solution. Strictly speaking, two solutions become coincident, that is x1=x2=-b/(2a);
In this case the left-hand side of the equation is a complete square.

Example: x2-4x+4=0; left hand side is simply (x-2)2; x1=x2=2;

3) D<0; There are no real solutions. There are two imaginary solutions, though.

Example:
x2-x+1=0; D=12-4*1*1=-3<0; No real solutions exist.

If two quadratic equations have identical solutions, then the equations are identical up to a factor.

Example:

x2-5x+6=0 and 3x2-15x+18=0; The second equation is just the first equation multiplied by 3.