Chapter 9.6, Problem 8CYU

To determine

Solve the equation.

Answer to Problem 8CYU

Quadratic Formula

Explanation of Solution

Given Information: $2x^2-3x-6=0$

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

Calculation:

• $D$ equal to negative so $x$ has no real roots.
• $D$ equal to positive so $x$ has real roots.

  $2x^2-3x-6=0$

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

As $a=2,b=-3$ and $c=-6$

  $D={\left(-3\right)}^2-4\times 2\times \left(-6\right)$

Now,

  $D=9+48$

Hence,

  $D=57$

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

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-3\right)\pm \sqrt{57}}{2\times 2}$

Now, square root of 57 is 7.55

  $x=\frac{3\pm 7.55}{2\times 2}$

Now,

$x=\frac{3+7.55}{4}$ or $x=\frac{3-7.55}{4}$

Now,

$x=\frac{10.55}{4}$ or $x=\frac{-4.55}{4}$

Now,

$x=2.64$ or $x=-1.14$

Nearest Tenth $x=2.6$ or $x=-1.1$