Solve the simultaneous equations y = √ $x$ $x$ +2 and ( $y + x$ $y+x$ )( $y - x$ $y-x$ )=0 ?

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Solve the simultaneous equations y = √ $x$ $x$ +2 and ( $y + x$ $y+x$ )( $y - x$ $y-x$ )=0 ?

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Answer 1 · 2017-11-18T12:15:49

$(- 1, 1), (2, 2)$ $(-1,1),(2,2)$

Explanation:

$y = \sqrt{x + 2} \to (1)$ $y=sqrt(x+2)to(1)$

$(y + x) (y - x) = 0 \leftarrow factors of difference of squares$ $(y+x)(y-x)=0larrcolor(blue)"factors of difference of squares"$

$\Rightarrow y^{2} - x^{2} = 0 \to (2)$ $rArry^2-x^2=0to(2)$

$substitute y = \sqrt{x + 2} into equation (2)$ $color(blue)"substitute "y=sqrt(x+2)" into equation "(2)$

${(\sqrt{x + 2})}^{2} - x^{2} = 0$ $(sqrt(x+2))^2-x^2=0$

$> \Rightarrow x + 2 - x^{2} = 0$ $>rArrx+2-x^2=0$

$multiply through by - 1$ $"multiply through by "-1$

$x^{2} - x - 2 = 0 \leftarrow in standard form$ $x^2-x-2=0larrcolor(blue)"in standard form"$

$the factors of - 2 which sum to - 1 are +1 and - 2$ $"the factors of - 2 which sum to - 1 are +1 and - 2"$

$\Rightarrow (x + 1) (x - 2) = 0$ $rArr(x+1)(x-2)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$x + 1 = 0 \Rightarrow x = - 1$ $x+1=0rArrx=-1$

$x - 2 = 0 \Rightarrow x = 2$ $x-2=0rArrx=2$

$substitute these values into equation (1)$ $"substitute these values into equation "(1)$

$x = - 1 \to y = \sqrt{- 1 + 2} = 1$ $x=-1toy=sqrt(-1+2)=1$

$x = 2 \to y = \sqrt{2 + 2} = 2$ $x=2toy=sqrt(2+2)=2$

$points of intersection are (- 1, 1) and (2, 2)$ $"points of intersection are "(-1,1)" and "(2,2)$

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Solve the simultaneous equations y = √ $x$ $x$ +2 and ( $y + x$ $y+x$ )( $y - x$ $y-x$ )=0 ?

1 Answer

Explanation:

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Solve the simultaneous equations y = √ $x$ $x$ +2 and ( $y + x$ $y+x$ )( $y - x$ $y-x$ )=0 ?