Chapter EP, Problem 5.6.14EP

(a)

To determine

To Find: The value of the discriminant.

Answer to Problem 5.6.14EP

  $16$

Explanation of Solution

Given:

  $4x^2+16x+15=0$

Formula Used:

  $D=b^2-4ac$

Calculation:

  $\begin{array}{l}4x^2+16x+15=0\\ a=4\\ b=16\\ c=15\\ D=256-4\times 4\times 15=256-240=16\end{array}$

Conclusion:

Thus the discriminant is $16$

(b)

To determine

To Find: The number and type of roots.

Answer to Problem 5.6.14EP

Two distinct and real

Explanation of Solution

Given:

  $4x^2+16x+15=0$

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:

  $\begin{array}{l}4x^2+16x+15=0\\ a=4\\ b=16\\ c=15\\ D=256-4\times 4\times 15=256-240=16>0\end{array}$

Conclusion:

Thus the roots are distinct and real.

(c)

To determine

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

Answer to Problem 5.6.14EP

  $-\frac{4}{5}or-\frac{3}{4}$

Explanation of Solution

Given:

  $4x^2+16x+15=0$

Formula Used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}4x^2+16x+15=0\\ a=4\\ b=16\\ c=15\\ D=256-4\times 4\times 15=256-240=16>0\\ x=\frac{-16\pm \sqrt{D}}{4\left(4\right)}=\frac{-16\pm 4}{16}\\ \Rightarrow x=-\frac{4}{5}or-\frac{3}{4}\end{array}$

Conclusion:

Thus the solution of the given quadratic equation is $-\frac{4}{5}or-\frac{3}{4}$