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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1a) To solve each of these equations by factorizing the left-hand side into a product of linear factors:
(i) x^2 - 6x - 40 = 0
To factorize this quadratic equation, we need to find two numbers whose product is -40 and whose sum is -6. In this case, -10 and 4 satisfy these conditions since -10 * 4 = -40 and -10 + 4 = -6.
Therefore, we can rewrite the equation as:
(x - 10)(x + 4) = 0
Setting each factor equal to zero and solving for x gives us the solutions:
x - 10 = 0 --> x = 10
x + 4 = 0 --> x = -4
So the solutions to this equation are x = 10 and x = -4.
(ii) 3x^2 + 20x - 7 = 0
To factorize this equation, we need to find two numbers whose product is -7 and whose sum is 20. In this case, 7 and -1 satisfy these conditions since 7 * (-1) = -7 and 7 + (-1) = 6.
Therefore, we can rewrite the equation as:
(3x + 7)(x - 1) = 0
Setting each factor equal to zero and solving for x gives us the solutions:
3x + 7 = 0 --> 3x = -7 --> x = -7/3
x - 1 = 0 --> x = 1
So the solutions to this equation are x = -7/3 and x = 1.
(b) To check the solutions by substituting them back into the equation:
For the first equation, x = 10:
x^2 - 6x - 40 = 0
(10)^2 - 6(10) - 40 = 0
100 - 60 - 40 = 0
0 = 0
For the first equation, x = -4:
x^2 - 6x - 40 = 0
(-4)^2 - 6(-4) - 40 = 0
16 + 24 - 40 = 0
0 = 0
For the second equation, x = -7/3:
3x^2 + 20x - 7 = 0
3(-7/3)^2 + 20(-7/3) - 7 = 0
49/3 - 140/3 - 7 = 0
49 - 140 - 21 = 0
-112 = 0
For the second equation, x = 1:
3x^2 + 20x - 7 = 0
3(1)^2 + 20(1) - 7 = 0
3 + 20 - 7 = 0
16 = 0
Since all the results are not equal to zero, it shows that the solutions are correct.
3. To 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 need to add and subtract the square of half the coefficient of x from the equation:
(x^2 - 6x + 9) - 9 = 40
(x - 3)^2 = 49
Taking the square root of both sides gives:
x - 3 = ±√49
x - 3 = ±7
Solving for x, we get:
x = 3 + 7 = 10
x = 3 - 7 = -4
So the solutions to this equation are x = 10 and x = -4.
(b) 3x^2 + 20x = 7
Following the same procedure as above, we have:
(3x^2 + 20x + 100) - 100 = 7
(3x + 10)^2 = 107
Taking the square root of both sides gives:
3x + 10 = ±√107
Solving for x, we get:
3x = -10 ± √107
x = (-10 ± √107) / 3
So the solutions to this equation are x = (-10 + √107) / 3 and x = (-10 - √107) / 3.
4. To solve the equations from Question 1 using the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For equation (i) x^2 - 6x - 40 = 0:
a = 1, b = -6, and c = -40
Plugging the values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4(1)(-40))) / (2(1))
x = (6 ± √(36 + 160)) / 2
x = (6 ± √196) / 2
x = (6 ± 14) / 2
Solving for x, we get:
x = (6 + 14) / 2 = 20 / 2 = 10
x = (6 - 14) / 2 = -8 / 2 = -4
So the solutions for equation (i) are x = 10 and x = -4.
For equation (ii) 3x^2 + 20x - 7 = 0:
a = 3, b = 20, and c = -7
Plugging the values into the quadratic formula, we get:
x = (-(20) ± √((20)^2 - 4(3)(-7))) / (2(3))
x = (-20 ± √(400 + 84)) / 6
x = (-20 ± √484) / 6
x = (-20 ± 22) / 6
Solving for x, we get:
x = (-20 + 22) / 6 = 2 / 6 = 1/3
x = (-20 - 22) / 6 = -42 / 6 = -7/3
So the solutions for equation (ii) are x = 1/3 and x = -7/3.
To solve each equation by factorizing the left-hand side into a product of linear factors, follow these steps:
(i) x^2 - 6x - 40 = 0
- To factorize this equation, we need to find two numbers that multiply to give -40 and add up to -6.
- By trial and error, we can find that -10 and +4 satisfy these conditions.
- So, the equation can be rewritten as (x - 10)(x + 4) = 0.
- Now, we set each factor equal to zero and solve for x:
x - 10 = 0 or x + 4 = 0
x = 10 or x = -4
(ii) 3x^2 + 20x - 7 = 0
- Again, we need to find two numbers that multiply to give -21 and add up to +20.
- By trial and error, we can find that +21 and -1 satisfy these conditions.
- So, the equation can be rewritten as (3x + 21)(x - 1) = 0.
- Setting each factor equal to zero, we get:
3x + 21 = 0 or x - 1 = 0
3x = -21 or x = 1
x = -7 or x = 1
To check the solutions obtained in part (a) by substituting them back into the equations:
(i) For x=10:
(10)^2 - 6(10) - 40 = 0
100 - 60 - 40 = 0
0 = 0 - The solution is verified.
(ii) For x=-4:
(-4)^2 - 6(-4) - 40 = 0
16 + 24 - 40 = 0
0 = 0 - The solution is verified.
(iii) For x=-7:
(3(-7))^2 + 20(-7) - 7 = 0
147 - 140 - 7 = 0
0 = 0 - The solution is verified.
(iv) For x=1:
(3(1))^2 + 20(1) - 7 = 0
3 + 20 - 7 = 0
0 = 0 - The solution is verified.
To solve each equation by completing the square on the left-hand side, follow these steps:
(a) x^2 - 6x = 40
- To complete the square, we need to add (b/2)^2 to both sides, where b is the coefficient of x.
- In this case, b = -6, so we add (-6/2)^2 = 9 to both sides:
x^2 - 6x + 9 = 40 + 9
(x - 3)^2 = 49
- Take the square root of both sides:
x - 3 = ± √49
- Solve for x by adding or subtracting the square root of 49:
x = 3 ± 7
x = 10 or x = -4
(b) 3x^2 + 20x = 7
- Again, we complete the square by adding (b/2)^2 to both sides, where b is the coefficient of x.
- In this case, b = 20, so we add (20/2)^2 = 100 to both sides:
3x^2 + 20x + 100 = 7 + 100
3(x^2 + (20/3)x + 100/3) = 107/3
- Divide both sides by 3 to isolate the squared term:
x^2 + (20/3)x + 100/3 = 107/9
- To complete the square, we need to add (b/2)^2 to both sides. Here, (20/3)/2 = 10/3, so we add (10/3)^2 = 100/9 to both sides:
x^2 + (20/3)x + 100/3 + 100/9 = 107/9 + 100/9
(x + 10/3)^2 = 207/9
- Take the square root of both sides:
x + 10/3 = ± √(207/9)
- Solve for x by adding or subtracting the square root of 207/9:
x = -10/3 ± √(207/9)
To solve the equations from Question 1 using the quadratic formula:
The quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
(i) For x^2 - 6x - 40 = 0:
- The coefficients are a = 1, b = -6, and c = -40.
- Substituting these values into the quadratic formula, we get:
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
(ii) For 3x^2 + 20x - 7 = 0:
- The coefficients are a = 3, b = 20, and c = -7.
- Substituting these values into the quadratic formula, we get:
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
These are the solutions to the equations from Question 1 using the quadratic formula.