(Δ=)b2−4ac is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if Δ<0 then the equation has complex roots (non real), if Δ=0 then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if Δ>0 then the equation has 2 distinct real roots. The general solution to the quadratic you gave is 2a−b±b2−4ac. So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the x-axis or not. If Δ<0 then the curve y does not cross the x-axis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis. For example if is b2−4ac=−9, what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.
b2 - 4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.
It's shown as part of the quadratic formula
where it's evident that it cannot be solved if the discriminant is less than zero.
It's actually derived by completing the square. For the equation ax2 + bx + c = 0:
x^2 + (b/a)x + c/a = 0
(x + b/2a)2- (b/2a)2 + c/a = 0
(x + b/2a)2 = (b/2a)2 - c/a
As the LHS is a square:
(b/2a)2 - c/a ≥0
(b2/4a2) -c/a ≥0
(b2/4a2) -4ac/4a2≥0
(b2 - 4ac)/4a2≥0
And therefore, as 4a2 is always positive, b2 - 4ac ≥0 for there to be real solutions of x.
(Δ=)b2−4ac is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if Δ<0 then the equation has complex roots (non real), if Δ=0 then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if Δ>0 then the equation has 2 distinct real roots. The general solution to the quadratic you gave is 2a−b±b2−4ac. So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the x-axis or not. If Δ<0 then the curve y does not cross the x-axis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis. For example if is b2−4ac=−9, what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.
b2 - 4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.
It's shown as part of the quadratic formula
where it's evident that it cannot be solved if the discriminant is less than zero.
It's actually derived by completing the square. For the equation ax2 + bx + c = 0:
x^2 + (b/a)x + c/a = 0
(x + b/2a)2- (b/2a)2 + c/a = 0
(x + b/2a)2 = (b/2a)2 - c/a
As the LHS is a square:
(b/2a)2 - c/a ≥0
(b2/4a2) -c/a ≥0
(b2/4a2) -4ac/4a2≥0
(b2 - 4ac)/4a2≥0
And therefore, as 4a2 is always positive, b2 - 4ac ≥0 for there to be real solutions of x.
Thank you very much. But question does not ask for equality. It asks for inequality.
Cool. Took the ratio of two formulas and ans is 1. Verified. I'm gonna use it where ever possible . :d
They are just really the same formula, just rationalise the denominator to get to the more conventional one. Advantage of this one though is that it gives you a solution even if a=0.
They are just really the same formula, just rationalise the denominator to get to the more conventional one. Advantage of this one though is that it gives you a solution even if a=0.
Ah. It falsely gives you two roots for linear equation. Though the second root is undefined, does it not violate one of the fundamental theorems of algebra? A nth degree equation has n-roots, no more.
Ah. It falsely gives you two roots for linear equation. Though the second root is undefined, does it not violate one of the fundamental theorems of algebra? A nth degree equation has n-roots, no more.
Anyway thanks.
You would have to take the positive root of the b^2 because otherwise you would get 0 on the denominator. It gives x=-c/b which of course is the solution to bx+c=0.