Respuesta :
y = x2 – 3x – 10
roots are x intercepts on form of (x,0)
Let's solve the quadratic equation
y = x2 – 3x – 10
(x-5)(x+2)
x=5,-2
roots: (5,0)(-2,0)
let's see of these are correct
y = x2 – 3x – 10
y=(5)^2-3 (5)-10
y=25-15-10
y=10-10
y=0
y = x2 – 3x – 10
y=(-2)^2-3 (-2)-10
y=4+6-10
y=10-10
y=0
the roots are correct
roots are x intercepts on form of (x,0)
Let's solve the quadratic equation
y = x2 – 3x – 10
(x-5)(x+2)
x=5,-2
roots: (5,0)(-2,0)
let's see of these are correct
y = x2 – 3x – 10
y=(5)^2-3 (5)-10
y=25-15-10
y=10-10
y=0
y = x2 – 3x – 10
y=(-2)^2-3 (-2)-10
y=4+6-10
y=10-10
y=0
the roots are correct
Quadratic equation can be written in the form ax²+bx+c. The roots of the quadratic equation [tex]y = x^2-3x-10[/tex]-2 and 5.
What is a Quadratic Equation?
A quadratic equation is an equation that can be written in the form of
ax²+bx+c.
Where a is the leading coefficient, and
c is the constant.
How to find replace the mid-value of a quadratic equation?
To find the roots of the given quadratic equation we will find two numbers whose sum is equal to -3 and the product is equal to -10 to replace -3 in the given quadratic equation.
Given to us[tex]y = x^2-3x-10[/tex]
The quadratic equation can be written as,
[tex]\begin{aligned}y&=x^2-3x-10\\&=x^2-5x+2x-10\end{aligned}[/tex]
Taking x common from the first two terms while 2 as common from the last two terms,
[tex]=x^2-5x+2x-10\\\\=x(x-5)+2(x-5)\\\\= (x+2)(x-5)[/tex]
equating the two factors we get against zero,
(x+2) = 0
x = -2
(x-5) =0
x =5
Hence, the factors of the given quadratic equation are -2 and 5.
Learn more about Quadratic equations:
https://brainly.com/question/2263981