Question

Составьте квадратное уравнение,корни которого равны:2+(кв. Корень из)3 и 2-(кв. Корень из)3 Помогите с решением!

Answer 1

X1=2-√3,x2=2+√3
p=-(x1+x2)=-(2-√3+2+√3)=-4
q=x1*x2=(2-√3)(2+√3)=4-3=1
x²-4x+1=0

Answer 2

Quadratic Equation with Given Roots

To find the quadratic equation with roots \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\), we can use the fact that if \(r\) and \(s\) are the roots of a quadratic equation \(ax^2 + bx + c = 0\), then the equation can be expressed as \((x - r)(x - s) = 0\).

So, the quadratic equation with the given roots is:

\((x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = 0\)

Expanding this equation gives us the quadratic equation.

Solution

Expanding the above expression:

\((x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = 0\)

\((x - 2 - \sqrt{3})(x - 2 + \sqrt{3}) = 0\)

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

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

So, the quadratic equation with the given roots \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\) is:

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