0.4286 would be the decimal
(x+3)(x+3) = x^2 -3 + 3x
x^2 + 3x + 3x + 9 = x^2 + 3x - 3
x^2 + 6x + 9 = x^2 + 3x - 3
6x + 9 = 3x - 3
3x + 9 = -3
3x = -12
x = -4
Check:
(-4+3)^2 = (-4)^2 - 3(1+4)
(-1)^2 = 16 - 3(5)
1 = 16-15
1 = 1 :)
Answer:
![\displaystyle 2 {x}^{2} + 4x - 30 = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%202%20%7Bx%7D%5E%7B2%7D%20%20%2B%204x%20-%2030%20%3D%200)
Step-by-step explanation:
we are given the zeros and a point where it goes through of a quadratic equation
remember that when the roots are given then the function should be
![\displaystyle \: y = a(x - x_{1})(x - x_{2})](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5C%3A%20y%20%3D%20%20a%28x%20-%20%20x_%7B1%7D%29%28x%20-%20%20x_%7B2%7D%29%20)
where a is the leading coefficient and x1 and x2 are the roots
substitute:
![\displaystyle y = a(x - (3))(x - ( - 5))](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%20%3D%20a%28x%20-%20%20%283%29%29%28x%20-%20%20%28%20-%205%29%29%20)
simplify:
![\displaystyle y = a(x - 3)(x + 5)](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%20%3D%20a%28x%20-%20%203%29%28x%20%20%2B%205%29%20)
now the given point tells us that when x is 2 y is -14 therefore by using the point we can figure out a
substitute:
![\displaystyle a(2 - 3)(2 + 5) = - 14](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20a%282%20-%20%203%29%282%20%20%2B%205%29%20%20%3D%20%20-%2014)
simplify parentheses:
![\displaystyle a( - 1)(7) = - 14](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20a%28%20%20-%201%29%287%29%20%20%3D%20%20-%2014)
simplify multiplication:
![\displaystyle - 7a = - 14](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20-%207a%20%3D%20%20-%2014)
divide both sides by -7:
![\displaystyle a = 2](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20a%20%3D%20%202)
altogether substitute:
![\displaystyle y = 2(x - 3)(x + 5)](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%20%3D%202%28x%20-%20%203%29%28x%20%20%2B%205%29%20)
since it want the equation y should be
![\displaystyle 2(x - 3)(x + 5) = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%202%28x%20-%20%203%29%28x%20%20%2B%205%29%20%20%3D%200)
recall quadratic equation standard form:
![\displaystyle {ax}^{2} + bx + c = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%20%7Bax%7D%5E%7B2%7D%20%20%2B%20bx%20%2B%20c%20%3D%200)
so simplify parentheses:
![\displaystyle 2( {x}^{2} + 2x - 15 ) = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%202%28%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%20-%2015%20%29%20%20%3D%200)
distribute:
![\displaystyle 2 {x}^{2} + 4x - 30 = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%202%20%7Bx%7D%5E%7B2%7D%20%20%2B%204x%20-%2030%20%3D%200)
hence,
the equation of the parabola in standard form is <u>2</u><u>x</u><u>²</u><u>+</u><u>4</u><u>x</u><u>-</u><u>3</u><u>0</u><u>=</u><u>0</u>
Answer:
.
Step-by-step explanation: