Chapter 9.6, Problem 7CYU

To determine

Solve the equation.

Answer to Problem 7CYU

Quadratic formula

Explanation of Solution

Given Information: $2x^2+11x-6=0$

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.

  $2x^2+11x-6=0$

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

As $a=2,b=11$ and $c=-6$

  $D={\left(11\right)}^2-4\times 2\times \left(-6\right)$

Now,

  $D=121+48$

Hence,

  $D=169$

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

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-11\pm \sqrt{169}}{2\times 2}$

Now, square root of 169 is 13

  $x=\frac{-11\pm 13}{4}$

Now,

$x=\frac{-11+13}{4}$ or $x=\frac{-11-13}{4}$

Now,

$x=\frac{2}{4}$ or $x=\frac{-24}{2}$

Now,

$x=0.5$ or $x=-12$