Chapter 9.6, Problem 10CYU

To determine

Solve the equation.

Answer to Problem 10CYU

Quadratic Formula

Explanation of Solution

Given Information: $x^2-9x=-19$

Formula Used: $D=b^2-4ac$ and $x=\frac{-b\pm \sqrt{D}}{2a}$

Calculation:

• $D$ equal to negative so $x$ has no real solution.
• $D$ equal to 0 so $x$ has 1 real solution.
• $D$ equal to positive so $x$ has 2 real solution.

  $x^2-9x=-19$

So, simplify

  $x^2-9x+19=0$

Now, use formula $D=b^2-4ac$ to check if solution of $x$ exist or not.

As $a=1,b=-9$ and $c=19$

  $D={\left(-9\right)}^2-4\times 1\times 19$

Now,

  $D=81-76$

Hence,

  $D=5$

As, $D$ equal to positive so $x$ has 2 real solution.

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-9\right)\pm \sqrt{5}}{2\times 1}$

Now, square root of 5 is 2.24

  $x=\frac{9\pm 2.24}{2}$

Now,

$x=\frac{9+2.24}{2}$ or $x=\frac{9-2.24}{2}$

Now,

$x=\frac{11.24}{2}$ or $x=\frac{6.76}{2}$

Now,

$x=5.62$ or $x=3.38$

Nearest Tenth $x=5.6$ or $x=3.4$