Select the two values of x that are roots of this equation. x^2+2x-4=0

[tex]x^2+2x-4=0\qquad\text{add 4 to both sides}\\\\x^2+2x=4\\\\x^2+2(x)(1)=4\qquad\text{add}\ 1^2=1\ \text{to both sides}\\\\\underbrace{x^2+2(x)(1)+1^2}=4+1^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+1)^2=5\to x+1=\pm\sqrt5\qquad\text{subtract 1 from both sides}\\\\x=-1\pm\sqrt5[/tex]

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shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-06-11T13:46:50+02:00

The two values of x for the given equation are ( -1 + √5 ) and ( - 1 - √5 ).

What is a quadratic equation?

The polynomial having a degree of two or the maximum power of the variable in a polynomial will be 2 is defined as the quadratic equation and it will cut two intercepts on the graph at the x-axis.

The solution of the given equation is:-

= x² + 2x - 4

x² + 2x = 4

x² + 2x ( 1 ) + 1² = 4 + 1²

( x + 1 )² = 5

x + 1 = ± √5

x = -1 ± √5

Therefore the two values of x for the given equation are ( -1 + √5 ) and ( - 1 - √5 ).

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