(1) using the formula methods, solbe these quadratic equation
1)6xx +13x +6=0
2)3xx +12x +10=0
(2) using the factorization method , solve these quadratic equation
1) pp + 7p + 12=0
2) 2xx =3x + 5
(1) Using the formula methods, solve these quadratic equations:
1) 6x^2 + 13x + 6 = 0
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 6, b = 13, and c = 6. Substituting these values into the quadratic formula:
x = (-13 ± √(13^2 - 4*6*6)) / (2*6)
x = (-13 ± √(169 - 144)) / 12
x = (-13 ± √25) / 12
x = (-13 ± 5) / 12
So the possible solutions are:
x1 = (-13 + 5) / 12 = -8/12 = -2/3
x2 = (-13 - 5) / 12 = -18/12 = -3/2
Therefore, the solutions for the equation 6x^2 + 13x + 6 = 0 are x = -2/3 and x = -3/2.
2) 3x^2 + 12x + 10 = 0
Using the quadratic formula with a = 3, b = 12, and c = 10:
x = (-12 ± √(12^2 - 4*3*10)) / (2*3)
x = (-12 ± √(144 - 120)) / 6
x = (-12 ± √24) / 6
x = (-12 ± 2√6) / 6
x = -2 ± √6 / 3
Therefore, the solutions for the equation 3x^2 + 12x + 10 = 0 are x = (-2 + √6) / 3 and x = (-2 - √6) / 3.
(2) Using the factorization method, solve these quadratic equations:
1) p^2 + 7p + 12 = 0
To factorize this equation, we need to find two numbers whose sum is 7 and whose product is 12. Those numbers are 3 and 4.
Therefore, the equation factors to:
(p + 3)(p + 4) = 0
Applying the zero product property, this equation can be solved by setting each factor equal to zero:
p + 3 = 0 or p + 4 = 0
Solving these equations gives:
p = -3 or p = -4
The solutions for the equation p^2 + 7p + 12 = 0 are p = -3 and p = -4.
2) 2x^2 = 3x + 5
First, we rearrange the equation to bring all terms to one side:
2x^2 - 3x - 5 = 0
This quadratic equation does not easily factorize, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 2, b = -3, and c = -5. Substituting these values into the quadratic formula:
x = (-(-3) ± √((-3)^2 - 4*2*(-5))) / (2*2)
x = (3 ± √(9 + 40)) / 4
x = (3 ± √49) / 4
x = (3 ± 7) / 4
The possible solutions are:
x1 = (3 + 7) / 4 = 10/4 = 5/2
x2 = (3 - 7) / 4 = -4/4 = -1
Therefore, the solutions for the equation 2x^2 = 3x + 5 are x = 5/2 and x = -1.
2x^2 - 3x - 5 = 0
( 2 x - 5 ) (x+1) = 0
Apologies for the mistake in my previous response. Let's correctly factorize the equation 2x^2 - 3x - 5 = 0.
To factorize this quadratic equation, we need to find two numbers whose product is the product of the coefficients of x^2 and the constant term, which is 2*(-5) = -10. We also need the sum of these two numbers to be the coefficient of x, which is -3.
After trying different combinations, we find that the numbers -5 and 2 satisfy these conditions.
Therefore, the equation factors to:
(2x - 5)(x + 1) = 0
Using the zero product property, we can set each factor equal to zero:
2x - 5 = 0 or x + 1 = 0
Solving these equations gives:
2x = 5 ---> x = 5/2
x = -1
Hence, the solutions for the equation 2x^2 - 3x - 5 = 0 are x = 5/2 and x = -1.