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
---------------------
2a
I assume you can apply the quadratic formula.
3x^2 + 20x - 7
The factors must be of the form
(3x?1)(x?7)
or
(3x?7)(x?1)
where the proper + or - sign must be used to get the +20x. Hint: 3*7 = 21
To solve by completing the square, proceed as in the examples in your text to get a completed square:
3x^2 + 20x = 7
3(x^2 + 20/3 x) = 7
Now add (10/3)^2 inside the parentheses to complete the square:
3(x^2 + 20/3 x + 100/9) = 7 + 3*100/9
3(x + 10/3)^2 = 121/3
(x + 10/3)^2 = 121/9 = (11/3)^2
x + 10/3 = ±11/3
x = 10/3 ± 11/3
x = 7 or -1/3
To solve each of these equations, we will go through each step and explain how to get the answer.
1a) Solve each equation by factorising the left-hand side into a product of linear factors:
(i) x^2 - 6x - 40 = 0
To factorize, we want to find two numbers that multiply to give -40 and add up to -6. In this case, -10 and +4 satisfy these conditions. So we can rewrite the equation as:
(x - 10)(x + 4) = 0
Now we can use the zero-product property which states that if a product of factors is equal to zero, then at least one of the factors must be zero. So we set each factor equal to zero:
x - 10 = 0 or x + 4 = 0
Solving these equations individually, we get:
x = 10 or x = -4
These are the solutions to the equation (i).
(ii) 3x^2 + 20x - 7 = 0
We want two numbers that multiply to give -7 and add up to +20. In this case, -1 and +7 satisfy these conditions. So we can rewrite the equation as:
(3x - 1)(x + 7) = 0
Setting each factor equal to zero:
3x - 1 = 0 or x + 7 = 0
Solving these equations individually, we get:
x = 1/3 or x = -7
These are the solutions to the equation (ii).
(b) For each question in part (a), check your solutions by substituting them back into the equation.
For equation (i), we substitute x = 10 and x = -4 into the equation:
(i) x^2 - 6x - 40 = 0
For x = 10: (10)^2 - 6(10) - 40 = 0
This expression equals zero, so the solution x = 10 is correct.
For x = -4: (-4)^2 - 6(-4) - 40 = 0
This expression also equals zero, so the solution x = -4 is correct.
Similarly, we can substitute the solutions into equation (ii) to verify them.
3. Solve each equation from Question 1 by completing the square on the left-hand side:
(a) x^2 - 6x = 40
To complete the square, we take half of the coefficient of x (-6/2 = -3) and square it (-3)^2 = 9. We add and subtract 9 to both sides of the equation:
x^2 - 6x + 9 - 9 = 40
(x - 3)^2 - 9 = 40
Now we can simplify this equation:
(x - 3)^2 = 49
Taking the square root of both sides, we get:
x - 3 = ±√49
Solving for x:
x - 3 = 7 or x - 3 = -7
x = 10 or x = -4
These are the solutions to equation (a).
(b) 3x^2 + 20x = 7
Following the same steps as above, we complete the square:
3x^2 + 20x = 7
Divide both sides by the coefficient of x^2 to make the coefficient 1:
x^2 + (20/3)x = 7/3
Now we complete the square on the left side:
x^2 + (20/3)x + (20/6)^2 - (20/6)^2 = 7/3
(x + (10/3))^2 - (100/36) = 7/3
Simplifying:
(x + (10/3))^2 = (7/3) + (100/36)
(x + (10/3))^2 = (84 +100)/36
(x + (10/3))^2 = 184/36
(x + (10/3))^2 = 23/9
Taking the square root:
x + (10/3) = ±√(23/9)
Solving for x:
x = -10/3 + √(23/9) or x = -10/3 - √(23/9)
These are the solutions to equation (b).
4. Solve the equations from Question 1 using the quadratic formula:
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using this formula, we can solve the equations from Question 1:
(i) x^2 - 6x - 40 = 0
a = 1, b = -6, c = -40
x = (-(-6) ± √((-6)^2 - 4(1)(-40))) / (2(1))
x = (6 ± √(36 + 160)) / 2
x = (6 ± √196) / 2
x = (6 ± 14) / 2
x = 20/2 or x = -8/2
x = 10 or x = -4
These are the same solutions we obtained earlier.
(ii) 3x^2 + 20x - 7 = 0
a = 3, b = 20, c = -7
x = (-(20) ± √((20)^2 - 4(3)(-7))) / (2(3))
x = (-20 ± √(400 + 84)) / 6
x = (-20 ± √484) / 6
x = (-20 ± 22) / 6
x = 2/6 or x = -42/6
x = 1/3 or x = -7
Once again, these are the same solutions we obtained earlier.
By following these steps and methods, we can solve each equation by factorization, completing the square, and using the quadratic formula, as well as check the solutions by substituting them back into the original equations.