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vfiekz [6]
3 years ago
7

Find the solutions to the following linear-quadratic systems algebraically. Select the ordered pair that is one of the correct s

olutions from among the choices below.
y = x² +3x + 8
y = 2x + 10

(-2, 6)
(0, 8)
(0,10)
(2, 14)
Mathematics
1 answer:
Ghella [55]3 years ago
3 0

Answer:

(-2, 6)

Step-by-step explanation:

we have

y=x^{2} +3x+8 ----> equation A

y=2x+10 ----> equation B

equate equation A and equation B

x^{2} +3x+8=2x+10

x^{2} +3x+8-2x-10=0

x^{2} +x-2=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2} +x-2=0

so

a=1\\b=1\\c=-2

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}

x=\frac{-1\pm\sqrt{9}} {2}

x=\frac{-1\pm3} {2}

x=\frac{-1+3} {2}=1

x=\frac{-1-3} {2}=-2

Find the values of y

For x=1

y=2(1)+10=12

For x=-2

y=2(-2)+10=6

therefore

The solutions are (1,12) and (-2,6)

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Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

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