1a) Solve the each of these equations by factorising the left hand side into a product of linear factors:
(i) x^2 - 6x - 40 = 0
(ii) 3x^2 + 20x - 7 = 0
(b) For each question in part (a), check your solutions by substituting them back into the equation.
3. Solve each equation from Question 1, in the form given below,by completing the square on the left-hand side:
(a) x^2 - 6x = 40
(b) 3x^2 + 20x = 7
4. Solve the equations from Question 1 using the quadratic formula:
x = -b +- √b^2-4ac
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2a
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To solve each of the equations in Question 1(a) by factorizing the left-hand side into a product of linear factors, you can follow these steps:
(i) x^2 - 6x - 40 = 0
In this equation, you need to find two linear factors which, when multiplied, give you the original equation.
Step 1: Split the middle term (-6x) into two terms whose coefficients multiply to give you the product of the coefficient of x^2 (1) and the constant term (-40). In this case, the two terms are -10x and +4x.
x^2 - 10x + 4x - 40 = 0
Step 2: Group the terms and factor by grouping:
(x^2 - 10x) + (4x - 40) = 0
x(x - 10) + 4(x - 10) = 0
(x + 4)(x - 10) = 0
Step 3: Set each factor equal to zero and solve for x:
x + 4 = 0 -> x = -4
x - 10 = 0 -> x = 10
Therefore, the solutions to the equation x^2 - 6x - 40 = 0 are x = -4 and x = 10.
(ii) 3x^2 + 20x - 7 = 0
Following the same steps as above:
Step 1: Split the middle term (20x) into two terms whose coefficients multiply to give you the product of the coefficient of x^2 (3) and the constant term (-7). In this case, the two terms are 21x and -x.
3x^2 + 21x - x - 7 = 0
Step 2: Group the terms and factor by grouping:
(3x^2 + 21x) + (-x - 7) = 0
3x(x + 7) - 1(x + 7) = 0
(3x - 1)(x + 7) = 0
Step 3: Set each factor equal to zero and solve for x:
3x - 1 = 0 -> x = 1/3
x + 7 = 0 -> x = -7
Therefore, the solutions to the equation 3x^2 + 20x - 7 = 0 are x = 1/3 and x = -7.
To check the solutions in part (b), substitute each solution back into the original equation and see if it satisfies the equation.
For example, for the equation x^2 - 6x - 40 = 0, we found two solutions: x = -4 and x = 10. Substitute each solution into the equation to check:
Substituting x = -4:
(-4)^2 - 6(-4) - 40 = 16 + 24 - 40 = 0
The equation is satisfied.
Substituting x = 10:
(10)^2 - 6(10) - 40 = 100 - 60 - 40 = 0
The equation is satisfied.
Now, moving on to Question 3(a) and (b), to solve each equation by completing the square on the left-hand side:
(a) x^2 - 6x = 40
Step 1: Add the square of half the coefficient of x to both sides of the equation:
x^2 - 6x + 9 = 40 + 9
(x - 3)^2 = 49
Step 2: Take the square root of both sides:
x - 3 = ±√49
Step 3: Solve for x:
x - 3 = ±7
x = 3 ± 7
Therefore, the solutions to the equation x^2 - 6x = 40 are x = 3 + 7 and x = 3 - 7, which simplifies to x = 10 and x = -4.
(b) 3x^2 + 20x = 7
Following the same steps as above:
Step 1: Add the square of half the coefficient of x to both sides of the equation:
3x^2 + 20x + 100 = 7 + 100
(x + 10)^2 = 107
Step 2: Take the square root of both sides:
x + 10 = ±√107
Step 3: Solve for x:
x + 10 = ±√107
x = -10 ± √107
Therefore, the solutions to the equation 3x^2 + 20x = 7 are x = -10 + √107 and x = -10 - √107.
Lastly, to solve the equations from Question 1 using the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac))/(2a)
For each equation, you can substitute the values of a, b, and c into the formula to find the solutions.
For example, for the equation x^2 - 6x - 40 = 0:
a = 1, b = -6, c = -40
x = (-(-6) ± √((-6)^2 - 4(1)(-40)))/(2(1))
Simplifying further, x = (6 ± √(36 + 160))/2
x = (6 ± √196)/2
x = (6 ± 14)/2
Therefore, the solutions to the equation x^2 - 6x - 40 = 0 using the quadratic formula are x = (6 + 14)/2 and x = (6 - 14)/2, which simplify to x = 10 and x = -4 (same solutions as obtained from factorization method).
Similarly, you can apply the quadratic formula to the equation 3x^2 + 20x - 7 = 0 to find the solutions.