Chapter 9.4, Problem 3P

a)

To determine

To evaluate: the solutions of $x^2+14x=-49$ .

Answer to Problem 3P

The solution of above system of equations is (3, 8) and (−1, 4).

Explanation of Solution

Given:

  $x^2+14x=-49$

Using the quadratic formula,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

From the equation, $x^2+14x=-49$ finding the value of a , b and c ,

  $\begin{array}{l}a=1\\ b=14\\ c=49\end{array}$

Using quadratic formula to solve above quadratic equation,

  $\begin{array}{c}x=\frac{-14\pm \sqrt{{14}^2-4\left(1\right)\left(49\right)}}{2\left(1\right)}\\ =\frac{-14\pm \sqrt{196-196}}{2}\\ =\frac{-14\pm 0}{2}\\ =-7\end{array}$

So, the solution of equation is x = −7.

b)

To determine

To explain: the reason why quadratic equations of the form $x^2+2ax+a^2=0$ or $x^2-2ax+a^2=0$ have only one real-number solution.

Explanation of Solution

Given:

  $x^2+14x=-49$

Using the discriminant formula,

  $d=\sqrt{b^2-4ac}$

From the equation, $x^2+2ax+a^2=0$ finding the value of a , b and c ,

  $\begin{array}{l}a=1\\ b=2a\\ c=a^2\end{array}$

Using quadratic formula to solve above quadratic equation,

  $\begin{array}{c}d=\sqrt{4a^2-4\left(1\right)\left(a^2\right)}\\ =\sqrt{4a^2-4a^2}\\ =0\end{array}$

It can be seen that the value of discriminant is zero.

It is known that when the value of discriminant is zero, then quadratic equation will have one real-number solution.

Similarly,

From the equation, $x^2-2ax+a^2=0$ finding the value of a , b and c ,

  $\begin{array}{l}a=1\\ b=-2a\\ c=a^2\end{array}$

Using quadratic formula to solve above quadratic equation,

  $\begin{array}{c}d=\sqrt{4a^2-4\left(1\right)\left(a^2\right)}\\ =\sqrt{4a^2-4a^2}\\ =0\end{array}$

It can be seen that the value of discriminant is zero.

It is known that when the value of discriminant is zero, then quadratic equation will have one real-number solution.