Answer:
The ordered pairs of [tex]p(x) = x^{2}-2\cdot x -35[/tex] are [tex](r_{1}, p(r_{1})) = (-5, 0)[/tex] and [tex](r_{2},p(r_{2})) = (7,0)[/tex], respectively.
Step-by-step explanation:
Let be [tex]p(x)[/tex] a quadratic expression whose two real zeros are real, a value of [tex]x[/tex] is a zero of the expression when [tex]p(x) = 0[/tex]. The factored form of the expression is represented by [tex]p(x) = (x-r_{1})\cdot (x-r_{2})[/tex], where [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are the zeroes of the polynomial.
If [tex]p(x) = x^{2}-2\cdot x -35[/tex], then the ordered pairs that represent the zeros are:
[tex](r_{1}, p(r_{1})) = (-5, 0)[/tex] and [tex](r_{2},p(r_{2})) = (7,0)[/tex]
Quadratic equation has a leading coefficient of the second degree. The ordered pairs of the quadratic equation are (-5,0) and (7,0).
A quadratic equation is an equation whose leading coefficient is of second degree also the equation has only one unknown while it has 3 unknown numbers.
It is written in the form of ax²+bx+c.
As the factors of the quadratic equation are given, therefore, equating the factors with zero in order to get the value of x.
[tex](x+5)=0\\x = -5[/tex]
[tex](x-7)=0\\x = 7[/tex]
Now, if we see the quadratic equation or the function it is written as,
[tex]y = x^2-2x-35 \\\\ x^2-2x-35 = 0[/tex]
Therefore, the value of y is 0.
Thus, the ordered pairs of the quadratic equation are (-5,0) and (7,0).
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