Question

State whether the following quadratic equations have two distinct real roots. Justify your answer.

$\left(vi\right){(x-\sqrt{2})}^2-2(x+1)=0$

Answer 1

Nature of roots.

The discriminants of a quadratic equation $a{x}^2+bx+c=0$ is given by:

$D=b^2-4ac$

• If $D>0$, the roots are distinct and real.
• If $D=0$, the roots are equal and real.
• If $D<0$, the roots will be imaginary.

The given equation ${(x-\sqrt{2})}^2-2(x+1)=0$ can be simplified as:

$$\begin{gathered} \therefore {(x-\sqrt{2})}^2-2(x+1)=0 \\ \Rightarrow x^2+2-2\sqrt{2}x-2x-2=0 \\ \Rightarrow x^2-(2\sqrt{2}+2)x=0 \end{gathered}$$

The discriminant of the given quadratic equation $x^2-(2\sqrt{2}+2)x=0$is computed as:

$$\begin{gathered} \therefore D={\left[-(2\sqrt{2}+2)\right]}^2-4·1·0 \\ \Rightarrow D={(2\sqrt{2}+2)}^2 \\ \Rightarrow D>0 \end{gathered}$$

Since the discriminant is greater than zero.

Hence, the roots are distinct and real.