Chapter 1.8, Problem 4GP

Solution

After evaluating the discriminant of the quadratic equation $2x^2+4x-4=0$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $48$ and it has two distinct real solutions $x=\left\{-1+\sqrt{3},-1-\sqrt{3}\right\}$ .

Given:

The quadratic equation is $2x^2+4x-4=0$ .

Concept used:

Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(1\right)$

And the term $b^2-4ac$ is called discriminant $\left(D\right)$ of the quadratic equation. Mathematically, discriminant $\left(D\right)=b^2-4ac$ and to find nature and number of roots of quadratic equation three cases arises.

  $\left(1\right)$ If discriminant $D>0$ , then the quadratic equation has two real and distinct solutions.

  $\left(2\right)$ If discriminant $D<0$ , then the quadratic equation has two imaginary distinct solutions which will be complex conjugate of each other.

  $\left(1\right)$ If discriminant $D=0$ , then the quadratic equation a unique real solution.

Calculation:

Evaluate discriminant $\left(D\right)$ of the quadratic equation $2x^2+4x-4=0$ .

  $\begin{array}{c}D={\left(4\right)}^2-4\left(2\right)\left(-4\right)\\ =16+32\\ =48\\ >0\end{array}$

Now discriminant $D=48$ is greater than zero, therefore the quadratic equation has two distinct real solution using clause $\left(1\right)$ in concept state above.

Now, the solutions for the quadratic equation $2x^2+4x-4=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(4\right)\pm \sqrt{{\left(4\right)}^2-4\left(2\right)\left(-4\right)}}{2\left(2\right)}\mathrm{......}\left(\because 2x^2+4x-4=0\right)\\ =\frac{-4\pm 4\sqrt{3}}{4}\\ =-1\pm \sqrt{3}\\ =\left\{-1+\sqrt{3},-1-\sqrt{3}\right\}\end{array}$

Conclusion:

The discriminant of the quadratic equation is $48$ and it has two distinct real solutions $x=\left\{-1+\sqrt{3},-1-\sqrt{3}\right\}$ .