i forgot how to solve these types of problems..
2x^2-18x=0, x^2=6x+7, 6y^2+5y-6=0
Also
SOLVE FOR X CORRECT TO 4 DECIMAL PLACES:
x^2+6x+2=0 2x+4=3x^2
For the first bit of the question, get all the terms over onto the left-hand side of the equation. For example, take the second one:
X²-6X-7=0
Usually you can factorize it easily into the following form: (X+A)(X+B). (There's no guarantee that it's an easy factorization, but if you're expected to solve it this way it usually is.)
If you multiply out those two brackets, you'll get X² + (A+B)X + AB.
So in this example, if you compare the corresponding terms from the two equations, A + B = -6 and AB = -7. So what are A and B? Usually you can spot it in a few seconds...
How about A = 1 and B = -7?
One down, two to go...
To solve these types of problems, you can use different techniques such as factoring, completing the square, or using the quadratic formula. I will guide you through each problem step by step using the appropriate method.
Problem 1:
2x^2 - 18x = 0 (quadratic equation)
Step 1: Factor out the common factor, if any.
2x(x - 9) = 0
Step 2: Set each factor equal to zero and solve for x.
2x = 0 or x - 9 = 0
x = 0 or x = 9
The solutions to the equation are x = 0 and x = 9.
Problem 2:
x^2 = 6x + 7 (quadratic equation)
Step 1: Rewrite the equation in standard form by moving all the terms to one side.
x^2 - 6x - 7 = 0
Step 2: Solve the equation by factoring or quadratic formula. In this case, let's use factoring.
(x - 7)(x + 1) = 0
Step 3: Set each factor equal to zero and solve for x.
x - 7 = 0 or x + 1 = 0
x = 7 or x = -1
The solutions to the equation are x = 7 and x = -1.
Problem 3:
6y^2 + 5y - 6 = 0 (quadratic equation)
Step 1: Factor the quadratic equation.
(3y - 2)(2y + 3) = 0
Step 2: Set each factor equal to zero and solve for y.
3y - 2 = 0 or 2y + 3 = 0
y = 2/3 or y = -3/2
The solutions to the equation are y = 2/3 and y = -3/2.
Now let's move on to solving equations with decimal places:
Problem 4:
x^2 + 6x + 2 = 0 (quadratic equation)
Step 1: Use the quadratic formula to solve for x.
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 1, b = 6, and c = 2.
x = (-(6) ± √((6)^2 - 4(1)(2)))/(2(1))
x = (-6 ± √(36 - 8))/(2)
x = (-6 ± √28)/(2)
x = (-6 ± 2√7)/(2)
x = -3 ± √7
To get the answer correct to 4 decimal places:
x ≈ -3 + 2.6458 ≈ -0.6458
x ≈ -3 - 2.6458 ≈ -5.6458
Problem 5:
2x + 4 = 3x^2 (quadratic equation)
Step 1: Rewrite the equation in standard form by moving all the terms to one side.
3x^2 - 2x - 4 = 0
Step 2: Solve the equation by factoring or quadratic formula. In this case, let's use the quadratic formula.
x = (-(-2) ± √((-2)^2 - 4(3)(-4)))/(2(3))
x = (2 ± √(4 + 48))/(6)
x = (2 ± √52)/(6)
x = (2 ± 2√13)/(6)
x = (1 ± √13)/(3)
To get the answer correct to 4 decimal places:
x ≈ (1 + 3.6056)/(3) ≈ 1.8685
x ≈ (1 - 3.6056)/(3) ≈ -0.8022
I hope this explanation helps you understand how to solve these types of problems. Let me know if you have any more questions!