Chapter 1.8, Problem 9GP

Solution

After evaluating the discriminant of the quadratic equation $3x-5x^2+1=6-7x$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $0$ and it has a unique real solution $x=1$ .

Given:

The quadratic equation is $3x-5x^2+1=6-7x$ which can be written as $5x^2-10x+5=0$ .

Concept used:

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

And the term $D=b^2-4ac$ is called discriminant $\left(D\right)$ of the quadratic equation. Now 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.

  $\left(3\right)$ If discriminant $D=0$ , then the quadratic equation a unique real solution.

Calculation:

Evaluate discriminant $\left(D\right)$ of the quadratic equation $5x^2-10x+5=0$ .

  $\begin{array}{c}D={\left(-10\right)}^2-4\left(5\right)\left(5\right)\\ =0\end{array}$

Now discriminant $D=0$ , therefore the quadratic equation has a unique real solution using clause $\left(3\right)$ in concept state above.

Now, the solution for the quadratic equation $5x^2-10x+5=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-10\right)\pm \sqrt{0}}{2\left(5\right)}\mathrm{......}\left(\begin{array}{l}\because 5x^2-10x+5=0\\ \text{ and }D=0\end{array}\right)\\ =\frac{10}{10}\\ =1\end{array}$

Conclusion:

The discriminant of the quadratic equation is $0$ and it has a unique real solution $x=1$ .