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.
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To solve these questions by factorizing the left-hand side into a product of linear factors, we can use the method of factorization.
(i) x^2 - 6x - 40 = 0:
To factorize this quadratic equation, we need to find two numbers that add up to -6 and multiply to -40. The numbers that satisfy this condition are -10 and 4.
Therefore, the factorization can be written as (x - 10)(x + 4) = 0.
Using the zero product property, we can set each factor equal to zero and solve for x.
Setting (x - 10) = 0, we get x = 10.
Setting (x + 4) = 0, we get x = -4.
To check our solutions, we substitute them back into the original equation:
When x = 10, (10)^2 - 6(10) - 40 = 100 - 60 - 40 = 0.
When x = -4, (-4)^2 - 6(-4) - 40 = 16 + 24 - 40 = 0.
(ii) 3x^2 + 20x - 7 = 0:
To factorize this quadratic equation, we need to find two numbers that add up to 20 and multiply to (3)(-7) = -21. The numbers that satisfy this condition are -1 and 21.
Therefore, the factorization can be written as (3x - 1)(x + 7) = 0.
Using the zero product property, we can set each factor equal to zero and solve for x.
Setting (3x - 1) = 0, we get 3x = 1, x = 1/3.
Setting (x + 7) = 0, we get x = -7.
To check our solutions, we substitute them back into the original equation:
When x = 1/3, 3(1/3)^2 + 20(1/3) - 7 = 1/3 + 20/3 - 7 = 1/3 + 20/3 - 21/3 = 0.
When x = -7, 3(-7)^2 + 20(-7) - 7 = 3(49) - 140 - 7 = 147 - 140 - 7 = 0.
Hence, we have solved and checked the solutions for the given quadratic equations.