1. 2(x+3)squared-6x+5=0
Answer= 2xsquared+6x+23=0
2. 3xsquared-6x+5=-2(x+1)-10
Answer= 3xsquared-4x+17=0
How would I solve from this point?
squared= ^2
1. 2(x+3)^2-6x+5 = 0
(x+3)^2 = x^2 + 6x + 9
2x^2 +12x + 18 - 6x +5 = 0
Combine terms.
2x^2 + 6x +17 = 0
Unless you have a typo, I don't agree with your answer.
2. 3x^2 - 6x + 5 = 2x + 2 - 10
Combine terms.
3x^2 - 8x + 2 = 0
Remember that, if you subtract/add from one side, you have to do the same thing to the other side of the equation.
To solve the given quadratic equations, you can follow these steps:
1. Start by simplifying both sides of the equation and combining like terms if necessary.
For the first equation, 2(x+3)^2 - 6x + 5 = 0, expand the square by applying the distributive property:
2(x^2 + 6x + 9) - 6x + 5 = 0
2x^2 + 12x + 18 - 6x + 5 = 0
Combine like terms:
2x^2 + 6x + 23 = 0
For the second equation, 3x^2 - 6x + 5 = -2(x+1) - 10, distribute -2 on the right side:
3x^2 - 6x + 5 = -2x - 2 - 10
Combine like terms:
3x^2 - 6x + 5 = -2x - 12
2. Next, move all terms to one side of the equation to obtain a quadratic equation in standard form.
For both equations:
2x^2 + 6x + 23 = 0
3x^2 - 6x + 5 + 2x + 12 = 0
Rearrange terms and combine like terms:
2x^2 + 6x + 23 = 0
3x^2 - 4x + 17 = 0
3. Use the quadratic formula to find the solutions.
The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the first equation, 2x^2 + 6x + 23 = 0, identify a = 2, b = 6, and c = 23. Substitute these values into the quadratic formula and simplify:
x = (-6 ± √(6^2 - 4 * 2 * 23)) / (2 * 2)
x = (-6 ± √(36 - 184)) / 4
x = (-6 ± √(-148)) / 4
Since the term inside the square root (√(-148)) is negative, the given equation does not have real solutions. It can only be expressed in terms of complex numbers.
For the second equation, 3x^2 - 4x + 17 = 0, identify a = 3, b = -4, and c = 17. Substitute these values into the quadratic formula and simplify:
x = (-(-4) ± √((-4)^2 - 4 * 3 * 17)) / (2 * 3)
x = (4 ± √(16 - 204)) / 6
x = (4 ± √(-188)) / 6
Similarly, since the term inside the square root (√(-188)) is negative, this equation also does not have real solutions. It can only be expressed in terms of complex numbers.
Therefore, both equations cannot be completely solved using real numbers.