Question

The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

Answer 1

$y=a(x-x_1)(x-x_2)\\\\x_1=-4;\ x_2=3\ therefore\\\\y=a(x+4)(x-3)\\\\y=a(x^2+x-12)\\\\\boxed{y=ax^2+ax-12a}\ where\ a\in\mathbb{R}-\{0\}$

Answer 2

Answer:

y=x²+x-12

Step-by-step explanation:

The quadratic standard form is given by this formula: ax² +bx +c =0.

However all we have is the zeros of this function.

1) So let's start by plugging into x the zeros. We will have two linear functions

$y=ax^{2} +bx+c\\\\ y=a(3)^{2} +3b+c\\ y=9a+3b+c \\ y=16a+4b+c$

2) Solving the system of linear equations, by the Addition Method:

$\left \{{{9a+3b+c=0} \atop {16a-4b+c=0}} \right.\\ \left \{{{9a+3b+c*(-1)=0} \atop {16a-4b+c=0}} \right\\ \left \{{{-9a-3b-c=0} \atop {16a-4b+c=0}} \right. \\ \\ 7a-7b=0\\ a=b\\ \\9+3+c=0\\ c=-12$

Therefore a=b and c=-12 and x=-4 and x=3

Since a=b, let's assume a=1 and b=1 and as c=-12 is the product of -4*3 hence we can rewrite this equation grouping it, and then test it:

$y=(x+4)(x-3)\\ y=x^{2} -3x+4x-12\\ y=x^{2} +x-12$ Standard form