Solve the quadratic equation x^2 + x - 6 = 0

There are 3 methods for solving quadratic equations;FactorisingThe quadratic equationCompleting the SquareUsing method 1,We observe that 2 and 3 are factors of 6 and that 3 - 2 = 1 so by factoring out we have (x + 3)(x - 2) = x^2 + x - 6Now we observe that (x + 3)(x - 2) = 0 so we know that x + 3 = 0 and x - 2 = 0 are solutions to our question,Rearranging gives us that x = - 3 and x = 2We can check this by substituting each of our answers into the question, i.e. (-3)^2 + (-3) - 6 = 0 and (2)^2 + (2) - 6 = 0Using method 2,Using the quadratic equation on our question where a = 1, b = 1 and c = - 6 we have,x = (-1 plus minus squareroot( 12 - 4(1)(- 6))/(2x1)Which gives us,x = (-1 + 5)/2 and x = (-1 - 5)/2which simplifies to x = 2 and x = - 3We can check this by substituting each of our answers into the question, i.e. (-3)^2 + (-3) - 6 = 0 and (2)^2 + (2) - 6 = 0Using method 3,By completing the square we find,(x+1/2)2 - 1/4 - 6 = 0 Which rearranges to,(x + 1/2)2 = 25/4Square rooting each side we find that,x + 1/2 = plus minus 5/2Which means our answers are x = 2 and x = -3We can check this by substituting each of our answers into the question, i.e. (-3)^2 + (-3) - 6 = 0 and (2)^2 + (2) - 6 = 0In practice we wouldn't use each method as this takes too long but if you have enough time then using more than one method is a good way at checking answers.

Answered by Thomas A. Maths tutor

2847 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

3n + 2 < 14, and 6n / (n ^2 + 5) >1. Find the values that n can take.


Differentiate (2a+3)^5/2 with respect to a


What is 64^(2/3)?


Sam has a bag of 10 coloured balls, four are red and six are blue. Sam then picks two balls from the bag. What is the probability that Sam gets: I) Two red balls II) One red and one blue III) One red THEN one blue in that order.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences