a) Solve for x: 2x² + 5x = 3
b) Solve for x and y: x + y = 4 and x² + y² + xy - 12 =0
c) Solve for x: 8
2x + 3 = ½x²
d) Factorise: 6x² - 7xy - 3y²
e) Factorise: 4x³ + 5x²y - 4xy² - 5y³
a) To solve for x in the equation 2x² + 5x = 3, we can rearrange the equation as follows:
2x² + 5x - 3 = 0.
To find the values of x, we can use factoring or the quadratic formula. In this case, let's use factoring:
The equation can be factored as (2x - 1)(x + 3) = 0.
Setting each factor equal to zero, we have:
2x - 1 = 0, which gives x = 1/2.
x + 3 = 0, which gives x = -3.
Therefore, the solutions for x are x = 1/2 and x = -3.
b) To solve for x and y in the system of equations x + y = 4 and x² + y² + xy - 12 = 0, we can use the method of substitution.
From the first equation, we can express y as y = 4 - x.
Substituting this value of y into the second equation, we get:
x² + (4 - x)² + x(4 - x) - 12 = 0.
Expanding and simplifying this equation, we have:
x² + 16 - 8x + x² + 4x² - 4x - 12 = 0.
Combining like terms, we get:
6x² - 12x + 4 = 0.
Dividing both sides of the equation by 2, we have:
3x² - 6x + 2 = 0.
Using the quadratic formula, x = (-b ± √(b² - 4ac))/(2a), where a = 3, b = -6, and c = 2.
Substituting these values into the quadratic formula, we get:
x = (-(-6) ± √((-6)² - 4 * 3 * 2))/(2 * 3).
Simplifying this equation, we have:
x = (6 ± √(36 - 24))/6.
x = (6 ± √12)/6.
Using further simplification, we get:
x = (6 ± 2√3)/6.
Simplifying the expression, we have:
x = 1 ± √3/3.
Therefore, the solutions for x are x = 1 + √3/3 and x = 1 - √3/3.
Substituting these values of x into the first equation x + y = 4, we can find the values of y:
For x = 1 + √3/3:
(1 + √3/3) + y = 4.
y = 4 - 1 - √3/3.
y = 3 - √3/3.
For x = 1 - √3/3:
(1 - √3/3) + y = 4.
y = 4 - 1 + √3/3.
y = 3 + √3/3.
Therefore, the solutions for x and y are:
(x, y) = (1 + √3/3, 3 - √3/3) and (x, y) = (1 - √3/3, 3 + √3/3).
c) To solve for x in the equation 8(2x + 3) = 1/2x², we can simplify the equation as follows:
16x + 24 = 1/2x².
Multiplying through by 2, we have:
32x + 48 = x².
Rearranging this equation, we get:
x² - 32x - 48 = 0.
To find the values of x, we can use factoring or the quadratic formula. In this case, let's use factoring:
The equation can be factored as (x - 8)(x + 6) = 0.
Setting each factor equal to zero, we have:
x - 8 = 0, which gives x = 8.
x + 6 = 0, which gives x = -6.
Therefore, the solutions for x are x = 8 and x = -6.
d) To factorize the expression 6x² - 7xy - 3y², we can look for pairs of numbers that multiply to give the product of 6 * -3, which is -18, and add up to the coefficient of the middle term, which is -7.
The pair of numbers that satisfy these conditions are -9 and 2, since -9 * 2 = -18 and -9 + 2 = -7.
Using these numbers, we can split the middle term of -7xy as follows:
6x² - 9xy + 2xy - 3y².
Now, we can factor the expression by grouping:
(6x² - 9xy) + (2xy - 3y²).
Taking out the common factor from the first pair of terms, we get:
3x(2x - 3y).
Taking out the common factor from the second pair of terms, we get:
y(2x - 3y).
Therefore, the factored form of 6x² - 7xy - 3y² is 3x(2x - 3y) + y(2x - 3y), which can also be written as (3x + y)(2x - 3y).
e) To factorize the expression 4x³ + 5x²y - 4xy² - 5y³, we can look for pairs of terms that have a common factor.
In this case, we can factor out x² from the first two terms and -y² from the last two terms:
4x²(x + (5y/4)) - 4y²(x + (5y/4)).
Now, we can see that (x + (5y/4)) is a common factor in both terms. Factoring it out, we get:
(x + (5y/4))(4x² - 4y²).
Further, we can factor out 4, giving us:
4(x + (5y/4))(x² - y²).
Finally, we can factor the difference of squares, x² - y², as (x + y)(x - y):
4(x + (5y/4))(x + y)(x - y).
Therefore, the factored form of 4x³ + 5x²y - 4xy² - 5y³ is 4(x + (5y/4))(x + y)(x - y).
a) To solve the equation 2x² + 5x = 3, we can follow these steps:
1) Rewrite the equation in the form of a quadratic equation: 2x² + 5x - 3 = 0
2) Solve the quadratic equation by factoring, completing the square, or using the quadratic formula.
3) If factoring is possible, try to factor the quadratic equation. In this case, it cannot be easily factored, so we will use the quadratic formula.
4) The quadratic formula states that x = (-b ± √(b² - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.
5) Substituting the values from the equation a = 2, b = 5, and c = -3 into the quadratic formula gives us:
x = (-5 ± √(5² - 4*2*(-3))) / (2*2)
Simplifying further:
x = (-5 ± √(25 + 24)) / 4
x = (-5 ± √(49)) / 4.
6) Taking the square root of 49 gives us two possible solutions:
x₁ = (-5 + 7) / 4
x₂ = (-5 - 7) / 4
7) Simplifying further, we get:
x₁ = 2/4 = 1/2
x₂ = -12/4 = -3.
Therefore, the solutions for x in the equation 2x² + 5x = 3 are x = 1/2 and x = -3.
b) To solve the system of equations x + y = 4 and x² + y² + xy - 12 = 0, we can follow these steps:
1) Rearrange the first equation to express one of the variables in terms of the other. In this case, it is easier to express x in terms of y: x = 4 - y.
2) Substitute the expression for x in the second equation: (4 - y)² + y² + (4 - y)y - 12 = 0.
3) Expand and simplify the equation: 16 - 8y + y² + y² + 4y - y² - 12 = 0.
4) Combine like terms: 16 - 8y + 4y - 12 = 0.
5) Simplify further: 4 - 4y = 0.
6) Solve for y: 4y = 4 → y = 1.
7) Substitute the value of y into the first equation to find x: x + 1 = 4 → x = 3.
Therefore, the solutions for x and y in the given system of equations are x = 3 and y = 1.
c) To solve the equation 8(2x + 3) = 1/2x² for x, we can follow these steps:
1) Distribute 8 to the terms inside the parenthesis: 16x + 24 = 1/2x².
2) Multiply all terms by 2 to remove the fraction: 32x + 48 = x².
3) Rearrange the equation to bring all terms to one side: x² - 32x - 48 = 0.
4) Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, we will use factoring.
5) The quadratic equation can be factored as (x - 8)(x + 6) = 0.
6) Set each factor equal to zero: x - 8 = 0 → x = 8 and x + 6 = 0 → x = -6.
Therefore, the solutions for x in the equation 8(2x + 3) = 1/2x² are x = 8 and x = -6.
d) To factorize the expression 6x² - 7xy - 3y², we can follow these steps:
1) Look for common factors among all the terms. In this case, there are no common factors.
2) Determine if the expression can be factored as a perfect square trinomial. In this case, it cannot.
3) Look for a pattern in the expression. In this case, we observe that the expression resembles a quadratic equation in the form of ax² + bx + c.
4) Use the quadratic formula to find the solutions for x in the quadratic equation 6x² - 7xy - 3y² = 0. The quadratic formula is x = (-b ± √(b² - 4ac)) / (2a), where a = 6, b = -7y, and c = -3y².
5) Substitute the values into the quadratic formula and simplify to get the solutions for x.
x = (7y ± √((7y)² - 4*6*(-3y²))) / (2*6)
x = (7y ± √(49y² + 72y²)) / 12
x = (7y ± √(121y²)) / 12
x = (7y ± 11y) / 12
6) Simplify further:
x₁ = (7y + 11y) / 12 = 18y / 12 = 3y / 2
x₂ = (7y - 11y) / 12 = -4y / 12 = -y / 3.
Therefore, the factored form of the expression 6x² - 7xy - 3y² is (3y / 2 - x)(x + y / 3).
e) To factorize the expression 4x³ + 5x²y - 4xy² - 5y³, we can follow these steps:
1) Look for common factors among all the terms. In this case, there are no common factors.
2) Determine if the expression can be factored by grouping. In this case, it cannot.
3) Look for a pattern or a common factor. In this case, we observe that each term contains x and y, as well as a coefficient of either 4 or 5.
4) Use the factoring by grouping method to factorize the expression. Group the terms into pairs based on their common factor and factor out the greatest common factor from each pair. Then, factor out the common factor between the pairs.
5) Group the terms as follows: (4x³ + 5x²y) - (4xy² + 5y³).
6) Factor out the greatest common factor from each pair: (x²(4x + 5y)) - (y²(4x + 5y)).
7) Notice that there is a common factor of (4x + 5y) present in both terms.
8) Common factor between the pairs: (4x + 5y).
9) Therefore, the factored form of the expression 4x³ + 5x²y - 4xy² - 5y³ is (4x + 5y)(x² - y²).