Find the Quadratic Equation Given the Roots -2 , 1+ square root of 7

$- 2$ , $1 + \sqrt{7}$

$x = - 2$ and $x = 1 + \sqrt{7}$ are the two real distinct solutions for the quadratic equation, which means that $x + 2$ and $x - 1 + \sqrt{7}$ are the factors of the quadratic equation.

$(x + 2) (x - 1 + \sqrt{7}) = 0$

Expand $(x + 2) (x - 1 + \sqrt{7})$ by multiplying each term in the first expression by each term in the second expression.

$x \cdot x + x \cdot - 1 + x \sqrt{7} + 2 x + 2 \cdot - 1 + 2 \sqrt{7} = 0$

Simplify terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 + x \sqrt{7} + 2 x + 2 \cdot - 1 + 2 \sqrt{7} = 0$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x + x \sqrt{7} + 2 x + 2 \cdot - 1 + 2 \sqrt{7} = 0$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x + x \sqrt{7} + 2 x + 2 \cdot - 1 + 2 \sqrt{7} = 0$

Multiply $2$ by $- 1$ .

$x^{2} - x + x \sqrt{7} + 2 x - 2 + 2 \sqrt{7} = 0$

Add $- x$ and $2 x$ .

$x^{2} + x \sqrt{7} + x - 2 + 2 \sqrt{7} = 0$

The standard quadratic equation using the given set of solutions ${- 2, 1 + \sqrt{7}}$ is $y = x^{2} + x \sqrt{7} + x - 2 + 2 \sqrt{7}$ .

$y = x^{2} + x \sqrt{7} + x - 2 + 2 \sqrt{7}$

Find the Quadratic Equation Given the Roots -2 , 1+ square root of 7

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