Chapter 9.6, Problem 62PFA

(A)

To determine

Solve the equation.

Answer to Problem 62PFA

  $x=2.4$ or $x=-2.1$

Explanation of Solution

Given Information: $3x^2-x-15=0$

Formula Used: $D=b^2-4ac$

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.

  $3x^2-x-15=0$

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

As $a=3,b=-1$ and $c=-15$

  $D={\left(-1\right)}^2-4\times 3\times \left(-15\right)$

Now,

  $D=1+180$

Hence,

  $D=181$

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

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-1\right)\pm \sqrt{181}}{2\times 3}$

Now, square root of 181 is 13.45

  $x=\frac{1\pm 13.45}{6}$

Now,

  $x=\frac{1+13.45}{6}$ or $x=\frac{1-13.45}{6}$

Now,

  $x=\frac{14.45}{6}$ or $x=\frac{-12.45}{6}$

Hence,

  $x=2.4$ or $x=-2.1$

(B)

To determine

Solve the equation.

Answer to Problem 62PFA

  $x=3$ or $x=-2.5$

Explanation of Solution

Given Information: $2x^2-x-15=0$

Formula Used: $D=b^2-4ac$

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.

  $2x^2-x-15=0$

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

As $a=2,b=-1$ and $c=-15$

  $D={\left(-1\right)}^2-4\times 2\times \left(-15\right)$

Now,

  $D=1+120$

Hence,

  $D=121$

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

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-1\right)\pm \sqrt{121}}{2\times 2}$

Now, square root of 121 is 11

  $x=\frac{1\pm 11}{4}$

Now,

  $x=\frac{1+11}{4}$ or $x=\frac{1-11}{4}$

Now,

  $x=\frac{12}{4}$ or $x=\frac{-10}{4}$

Hence,

  $x=3$ or $x=-2.5$

(C)

To determine

Solve the equation.

Answer to Problem 62PFA

No real solution

Explanation of Solution

Given Information: $3x^2-4x+2=0$

Formula Used: $D=b^2-4ac$

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.

  $3x^2-4x+2=0$

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

As $a=3,b=-4$ and $c=2$

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

Now,

  $D=16-24$

Hence,

  $D=-8$

As, $D$ equal to negative so $x$ has no real solution.

(D)

To determine

Solve the equation.

Answer to Problem 62PFA

2 real solution

Explanation of Solution

Given Information: $2x^2-4x+2=0$

Formula Used: $D=b^2-4ac$

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.

  $2x^2-4x+2=0$

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

As $a=2,b=-4$ and $c=2$

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

Now,

  $D=16-16$

Hence,

  $D=0$

As, $D$ equal to 0 so $x$ has 1 real solution.

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-4\right)\pm \sqrt{0}}{2\times 2}$

Now, square root of 0 is 0

  $x=\frac{4\pm \sqrt{0}}{4}$

Now,

  $x=\frac{4}{4}$

Hence,

  $x=1$