Mathematics: Y=-x^2+2x+10 y=x+2 substitution need asap

Y=-x^2+2x+10 y=x+2 substitution need asap - id221024313897, weckesserj9492

EricaLSH3588

13.01.2022

$x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}$

Explanation:

You must find the solution of the system using the substitution method.

System:

$y=-x^2+2x+10\\\\ y=x+2$

Substitute x+2 for y:

x+2=y=-x^2+2x+10

Pass all the terms to one side:

$x^2-2x+x+2-10\\ \\ x^2-x-8=0$

Use the quadratic formula to solve:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}$

$x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}$

Use y = x + 2 to find the value of y:

$y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}$

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carnations

17.10.2021

the answer is 2√2.

Step-by-step explanation:

5(8-8x^2)=8

or,40-40x^2=8

or,40-8=40x^2

or,32=40x^2

or,32/40=x^2

or,4/5=x^2

or,√4/5= x

therefore.x=√4/5

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