Find the value of x: 2x^2 - 3x - 4 = 1

2x^2 - 3x - 4 = 1
To solve, we should bring all the terms to one side, and equate the equation to 0. To do this, we should subtract 1 from each side.
2x^2 - 3x - 4 - 1 = 1 - 12x^2 - 3x - 5 = 0
Next, we should factorise the equation to get two factors, in the form of (ax + b)(cx + d) = 0. To find 'a' and 'c', we should find two numbers that multiply together to make 2. In this case, 'a' and 'c' can only be 2 and 1. Thus, the factorised equation currently stands as such:
(2x + b)(x + d) = 0
In order to find 'b' and 'd', we should find two numbers that multiply together to make -5. In this case, 'b' and 'd' can either be 1 and -5, or 5 and -1. To find which is the solution, we put one set of values to the equation, use the FOIL method to multiply it out, and see if the result is equal to the original equation. Lets take 1 and -5.
(2x + 1)(x - 5) = 02x^2 - 10x + x - 5 = 02x^2 - 9x - 5 = 0
This solution is incorrect. Lets now try switching the numbers around:
(2x - 5)(x+1) = 02x^2 + 2x - 5x - 5 = 02x^2 - 3x - 5 = 0
This solution is correct, so we now know that 'b' = -5, and 'd' = 1. We can now solve for the values of x.
(2x - 5)(x + 1) = 0
As the equation equals 0, we can tell that one of the two factors must equal 0. Thus:
2x - 5 = 0 OR x + 1 = 0
Taking the first factor: Taking the second factor:
2x - 5 = 0 x + 1 = 02x = 5 x = -1x = 5/2

PS
Answered by Prithvie S. Maths tutor

4029 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How to solve Simultaneous Linear Equations, e.g. (4x + 5y = 17) and (3x + 2y = 4)


How do I solve simultaneous equations?


Solve algebraically the simultaneous equations: x^2 + 3x + 10 = y and; 4 - 2x = y. Give answers as co-ordinates.


simultaneous equations - 2x + y = 7 and 3x -y = 8


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning