1. The stright line y=x+3 cuts the curve x^2 + y^2 = 29 at two points.
a) Prove that x^2 + 3x - 10 = 0
b) Hence find the coordinates of the points.

First question

\( line:~~~~y=x+3 \)

\( curve:~~~~x^2+y^2=29 \)

now we have to replace the y of curve by the y of line, therefore

\( x^2+(x+3)^2=29 \)

\( x^2+x^2+6x+9=29 \)

\( 2x^2+6x+9-29=0 \)

\( 2x^2+6x-20=0 \)

we can multiply each member by \( \frac{1}{2} \)

\( \boxed{\boxed{x^2+3x-10=0}} \)

Now we have to find the roots of this funtion

\( x^2+3x-10=0 \)

Sum and produc or Bhaskara

Then we find two axis

\( \boxed{x_1=2~~and~~x_2=-5} \)

now we replace this values to find the Y, we can replace in curve or line equation, I’ll prefer to replace it in line equation.

\( y=x+3 \)

\( y_1=x_1+3 \)

\( y_1=2+3 \)

\( \boxed{y_1=5} \)

\( y_2=x_2+3 \)

\( y_2=-5+3 \)

\( \boxed{y_2=-2} \)

Therefore

\( \boxed{\boxed{P_1(2,5)~~and~~P_2(-5,2)}} \)
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The second question give to us

\( y=ax+b \)

\( P_1(2,13) \)

\( P_2(-1,11) \)

We just have to replace the value then we’ll get a linear system.

point 1

\( 13=2a+b \)

point 2

\( -11=-a+b \)

then our linear system will be

\( \begin{matrix}2a+b&=&13\\-a+b&=&-11\end{matrix} \)

I’ll multiply the second line by -1 and I’ll add to first one

\( \begin{matrix}2a+(a)+b+(-b)&=&13+(-11)\\-a+b&=&-11\end{matrix} \)

\( \begin{matrix}3a&=&24\\-a+b&=&-11\end{matrix} \)

\( \begin{matrix}a&=&8\\-a+b&=&-11\end{matrix} \)

therefore we can replace the value of a, at second line

\( \begin{matrix}a&=&8\\-8+b&=&-11\end{matrix} \)

\( \begin{matrix}a&=&8\\b&=&-11+8\end{matrix} \)

\( \boxed{\boxed{\begin{matrix}a&=&8\\b&=&-3\end{matrix}}} \)

then our function will be

\( \boxed{\boxed{y=8x-3}} \)

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The third one, we have

\( line:~~~~y=3x-4 \)

\( curve:~~~~y=x^2-2x-4 \)

This resolution will be the same of our first question.

Let’s replace the y of curve by the y of line

\( 3x-4=x^2-2x-4 \)

\( 0=x^2-2x-4-3x+4 \)

therefore

\( \boxed{\boxed{x^2-5x=0}} \) 

now we have to find the roots of this function.

\( x^2-5x=0 \)

put x in evidence

\( x*(x-5)=0 \)

\( \boxed{x_1=0~~and~~x_2=5} \)

then

\( y=3x-4 \)

\( y_1=3x_1-4 \)

\( y_1=3*0-4 \)

\( \boxed{y_1=-4} \)

\( y_2=3x_2-4 \)

\( y_2=3*5-4 \)

\( y_2=15-4 \)

\( \boxed{y_2=11} \)

our points will be

\( \boxed{\boxed{P_1(0,4)~~and~~P_2(5,11)}} \)

;)


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