Loading the lesson
Loading the lesson
Search AllRounder.ai
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Learn it your way
A short class with Sarah and her students. The other ways are under More.
Sign up free to listen. Transcripts are free to read.
For 2x² − 4x − 6 = 0, what are a, b and c?
a is 2, b is 4 and c is 6.
Look at the signs in front of each term.
Oh, b is −4 and c is −6, because the terms are minus 4x and minus 6.
Exactly. Most mistakes with the formula come from dropping a sign at this step.
So I should write a = 2, b = −4, c = −6 before doing anything else.
Work out b² − 4ac.
(−4)² is 16, and 4 times 2 times −6 is −48. 16 minus −48 is 64.
64 is positive and a perfect square. What does that tell us?
Two real solutions, and they will be whole numbers or fractions.
Now finish it: x = (4 ± 8) / 4.
12 over 4 is 3, and −4 over 4 is −1. So x = 3 or x = −1.
And dividing the equation by 2 gives (x − 3)(x + 1) = 0, which agrees.
A ball's height is h = −5t² + 20t + 2. When does it hit the ground?
When h = 0. So a = −5, b = 20, c = 2, and D = 400 + 40 = 440.
t = (−20 ± √440) / (−10). That gives about −0.0976 or 4.10.
Do we keep both?
No, time can't be negative, so the ball lands after about 4.10 seconds.
Good. Always check whether each solution makes sense in the real situation.
x = (−b ± √(b² − 4ac)) / (2a) solves every quadratic equation ax² + bx + c = 0.
read a, b and c with their signs; in 2x² − 4x − 6 = 0, b = −4 and c = −6.
D = b² − 4ac gives two real roots if positive, one repeated root if zero, none if negative.
if D is a perfect square, the roots are rational and the quadratic factorises.
in real problems, reject solutions that make no sense, such as a negative time.
Solve 2x² − 4x − 6 = 0: a = 2, b = −4, c = −6, D = 16 + 48 = 64, so x = (4 ± 8)/4, giving x = 3 or x = −1.
Solve x² + 3x − 5 = 0: D = 9 + 20 = 29, so x = (−3 ± √29)/2, about 1.19 or −4.19.
Solve −x² + 2x + 4 = 0: a = −1, b = 2, c = 4, D = 20, so x = (−2 ± √20)/(−2) = 1 ∓ √5, about 3.24 or −1.24.
A ball with height h = −5t² + 20t + 2 lands when h = 0: D = 440 and t = (−20 ± √440)/(−10), so t ≈ 4.10 s (the negative root is rejected).
x equals negative b, plus or minus the square root, of b squared minus four a c, all over two a, to boot.
A careful baker always weighs the ingredients (a, b and c, with their signs) before mixing. She checks the oven (the discriminant) to see whether the cake will come out as two layers, one, or none at all.
"Signs, D, Substitute, Split" - write the signs of a, b, c; find D; substitute; split the ± into two answers.
D.T.N. - D positive gives Two roots, D zero gives one (Touching), D negative gives None.
Term
State the quadratic formula.
Click to flip
Term
What are a, b and c in 2x² − 4x − 6 = 0?
Click to flip
Term
What is the discriminant of 2x² − 4x − 6?
Click to flip
Term
How many real solutions are there when D < 0?
Click to flip
Term
What are the solutions of 2x² − 4x − 6 = 0?
Click to flip
x = (−b ± √(b² − 4ac)) / (2a), which solves any equation ax² + bx + c = 0 with a ≠ 0.
The number multiplying a term, taken with its sign; in 2x² − 4x − 6, the coefficient of x is −4.
D = b² − 4ac, the expression under the square root in the quadratic formula.
A single solution that occurs when D = 0; the parabola touches the x-axis at its vertex.
A solution that can be written as a fraction; it occurs when D is a perfect square.
A solution involving a surd such as √29; it occurs when D is positive but not a perfect square.
Example 3 solves 2x² − 4x − 6 = 0 with the quadratic formula: with a = 2, b = −4 and c = −6 the discriminant is 64, so x = (4 ± 8)/4, giving x = 3 or x = −1. The formula works for every quadratic equation.
The quadratic formula, x = (−b ± √(b² − 4ac)) / (2a), solves any equation ax² + bx + c = 0, whether or not it factorises. In Example 3, 2x² − 4x − 6 = 0, we substitute a = 2, b = −4 and c = −6, work out the discriminant b² − 4ac = 16 + 48 = 64, and get x = (4 ± 8)/4, so x = 3 or x = −1. This lesson works through the example carefully, shows how the discriminant predicts the number of real solutions, and applies the formula to irrational roots, a negative leading coefficient and a real-world height problem, pointing out the sign errors that cause most mistakes.
This lesson is the third of three worked examples on solving quadratic equations. Example 1 used factoring and Example 2 completing the square. The quadratic formula comes from completing the square on the general equation ax² + bx + c = 0, so it always works.
The formula. For ax² + bx + c = 0 with a ≠ 0:
x = (−b ± √(b² − 4ac)) / (2a)
The expression under the square root, D = b² − 4ac, is called the discriminant.
Example 3: Solve 2x² − 4x − 6 = 0.
Checking. 2(3)² − 4(3) − 6 = 18 − 12 − 6 = 0 and 2(−1)² − 4(−1) − 6 = 2 + 4 − 6 = 0. You can also divide the equation by 2 to get x² − 2x − 3 = 0, which factorises as (x − 3)(x + 1) = 0. The formula and factoring agree, as they must.
What the discriminant tells you.
More worked examples.
Avoiding the usual errors.