Algebra
Grade 6-8The Quadratic Formula
The quadratic formula solves any quadratic equation of the form ax² + bx + c = 0, including ones that cannot be factorised, by substituting the values of a, b and c directly into a fixed formula. This lesson covers using the quadratic formula, rounding solutions to a given accuracy, using the discriminant to find the number of real solutions, and forming and solving a quadratic equation from a worded problem.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Using the quadratic formula
For an equation of the form ax² + bx + c = 0, substitute the values of a, b and c into x = (−b ± √(b² − 4ac)) ÷ 2a, then work out both solutions, one using + and one using −.
To solve 2x² + 3x − 7 = 0, here a = 2, b = 3 and c = −7. Substituting gives x = (−3 ± √(3² − 4(2)(−7))) ÷ 2(2) = (−3 ± √65) ÷ 4, so x ≈ 1.27 or x ≈ −2.77.
Rounding solutions to a given accuracy
Keep the full, unrounded value throughout the calculation, and only round the final answer to the accuracy stated in the question, such as decimal places or significant figures.
Solving 2x² + 5x − 9 = 0 with the quadratic formula gives exact solutions that need rounding to 3 significant figures, using the unrounded value under the square root right up until the last step.
Using the discriminant
The discriminant is the part of the formula under the square root, b² − 4ac. If it is positive, the equation has two real solutions; if it is zero, the equation has one repeated real solution; if it is negative, the equation has no real solutions.
For 2x² + 3x + 5 = 0, the discriminant is 3² − 4(2)(5) = 9 − 40 = −31, which is negative, so the equation has no real solutions.
Forming and solving a quadratic equation from a worded problem
Translate the information given into an equation, rearrange it into the form ax² + bx + c = 0, then solve using the quadratic formula and reject any solution that does not make sense in context, such as a negative length.
A rectangle has length (x + 4) cm, width x cm, and area 30 cm². This gives x(x + 4) = 30, which rearranges to x² + 4x − 30 = 0. Solving gives x ≈ 3.83 or x ≈ −7.83, and the negative solution is rejected since x must be positive.
Worked Examples
Three exam-style questions, fully solved.
Solve 2x² + 3x − 7 = 0. Give your answers correct to 2 decimal places.
Easy- 1.Identify the values: a = 2, b = 3, c = −7
- 2.Substitute into the formula: x = (−3 ± √(3² − 4(2)(−7))) ÷ 2(2) = (−3 ± √65) ÷ 4
Answer: x = 1.27 or x = −2.77
Work out the discriminant of 2x² + 3x + 5 = 0 and state the number of real solutions.
Medium- 1.Identify the values: a = 2, b = 3, c = 5
- 2.Substitute into b² − 4ac: 3² − 4(2)(5) = 9 − 40
Answer: −31, which is negative, so there are no real solutions
A rectangle has length (x + 4) cm and width x cm. Its area is 30 cm². Form and solve a quadratic equation to find the value of x, correct to 2 decimal places (x > 0).
Hard- 1.Form the equation from the area: x(x + 4) = 30
- 2.Rearrange into the form ax² + bx + c = 0: x² + 4x − 30 = 0
- 3.Solve using the quadratic formula: x = (−4 ± √(4² − 4(1)(−30))) ÷ 2(1), giving x = 3.83 or x = −7.83
- 4.Reject the negative solution, since x must be greater than 0
Answer: x = 3.83
Avoid These
The most common mistakes students make.
Substituting a value with the wrong sign into the formula, especially when b or c is negative.
Making an arithmetic slip working out b² − 4ac, especially forgetting that squaring a negative number gives a positive result.
Forgetting the ± sign in the formula, and only finding one of the two solutions instead of both.
Rounding too early during the calculation, instead of keeping the full value and only rounding the final answer to the accuracy requested.
In a worded problem, forgetting to reject a solution that does not make sense in context, such as a negative length.
FAQ
Questions parents and students ask.
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