Chapter EP, Problem 5.6.8EP

(a)

To determine

To Find: The value of the discriminant.

Answer to Problem 5.6.8EP

  $0$

Explanation of Solution

Given:

  $3x^2-30x+75=0$

Formula Used:

  $D=b^2-4ac$

Calculation:

  $\begin{array}{l}3x^2-30x+75=0\\ a=3\\ b=-30\\ c=75\\ D=900-4\times 3\times 75=0\end{array}$

Conclusion:

Thus the discriminant is $0$

(b)

To determine

To Find: The number and type of roots.

Answer to Problem 5.6.8EP

Both are real and equal

Explanation of Solution

Given:

  $3x^2-30x+75=0$

Formula Used:

  $\begin{array}{l}D=b^2-4ac\\ D>0\text{Two distinct real roots}\\ D=0\text{Both roots are real and equal}\\ D<0\text{Imaginary}\end{array}$

Calculation:

  $\begin{array}{l}3x^2-30x+75=0\\ a=3\\ b=-30\\ c=75\\ D=900-4\times 3\times 75=0\end{array}$

Conclusion:

Thus the roots are real and equal.

(c)

To determine

To Find: The solution of the given quadratic equation using quadratic formula.

Answer to Problem 5.6.8EP

  $5$

Explanation of Solution

Given:

  $3x^2-30x+75=0$

Formula Used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}3x^2-30x+75=0\\ a=3\\ b=-30\\ c=75\\ D=900-4\times 3\times 75=0\\ x=\frac{30\pm \sqrt{D}}{2\left(3\right)}=\frac{30}{6}\\ \Rightarrow x=5\end{array}$

Conclusion:

Thus the solution of the given quadratic equation is $5$