1a) Solve each of these questions by factorising the left hand side into a product of linear factors:
(i) x^2 - 6x - 40 = 0
(ii) 3x^2 + 20x - 7 = 0
For each question in part (a), check your solutions by substituting them back in to the equation.
If the coefficient of x^2 is 1, look for 2 numbers which when multiplied give you the constant, and which when added give you the coefficient of the middle term
that is,
(-10)(4) = -40
(-10) + 4 = -6
so (x - 10)(x + 4)
for the second one: 3x^2 + 20x - 7 = 0 ,
what method have you learned.
I hesitate to give you a solution using a method you do not know.
Have you learned decomposition?
No I haven't
To solve each of these quadratic equations by factorizing, we need to find two binomials that, when multiplied together, will give us the left-hand side of the equation.
(i) x^2 - 6x - 40 = 0:
To factorize this equation, we need to find two numbers that add up to -6 and multiply to -40. These numbers are 10 and -4.
So, we can rewrite the equation as:
(x + 10)(x - 4) = 0
Now, we have two possibilities:
x + 10 = 0, which gives x = -10
x - 4 = 0, which gives x = 4
To check our solutions, we substitute them back into the original equation:
For x = -10: (-10)^2 - 6(-10) - 40 = 100 + 60 - 40 = 120 - 40 = 80, which is not equal to 0.
For x = 4: 4^2 - 6(4) - 40 = 16 - 24 - 40 = -8 - 40 = -48, which is also not equal to 0.
Therefore, there are no solutions to this equation.
(ii) 3x^2 + 20x - 7 = 0:
To factorize this equation, we need to find two numbers that add up to 20 and multiply to -7. These numbers are 7 and -1.
So, we can rewrite the equation as:
(3x - 1)(x + 7) = 0
Now, we have two possibilities:
3x - 1 = 0, which gives x = 1/3
x + 7 = 0, which gives x = -7
To check our solutions, we substitute them back into the original equation:
For x = 1/3: 3(1/3)^2 + 20(1/3) - 7 = 1/3 + 20/3 - 7 = 21/3 - 7 = 7 - 7 = 0
For x = -7: 3(-7)^2 + 20(-7) - 7 = 3(49) - 140 - 7 = 147 - 140 - 7 = 0
Therefore, the solutions to this equation are x = 1/3 and x = -7.