Chapter 9.5, Problem 41PPS

To determine

( $2x^2-11x+3$)

Solve the equation.

Answer to Problem 41PPS

  $D=97$ and 2 real roots

Explanation of Solution

Given Information: $2x^2-11x+3=0$

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.

  $2x^2-11x+3=0$

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

As $a=2,b=-11$ and $c=3$

  $D={\left(-11\right)}^2-4\times 2\times \left(3\right)$

Now,

  $D=121-24$

Hence,

  $D=97$

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

To determine

( $3x^2+6x+9$)

Solve the equation.

Answer to Problem 41PPS

  $D=-72$ and No real roots

Explanation of Solution

Given Information: $3x^2+6x+9=0$

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.

  $3x^2+6x+9=0$

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

As $a=3,b=6$ and $c=9$

  $D={\left(6\right)}^2-4\times 3\times \left(9\right)$

Now,

  $D=36-108$

Hence,

  $D=-72$

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

To determine

( $x^2-2x+7$)

Solve the equation.

Answer to Problem 41PPS

  $D=-24$ and NO real roots

Explanation of Solution

Given Information: $x^2-2x+7=0$

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.

  $x^2-2x+7=0$

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

As $a=1,b=-2$ and $c=7$

  $D={\left(-2\right)}^2-4\times 1\times 7$

Now,

  $D=4-28$

Hence,

  $D=-24$

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

To determine

( $x^2+10x+25$)

Solve the equation.

Answer to Problem 41PPS

  $D=0$ and 1 real roots

Explanation of Solution

Given Information: $x^2+10x+25=0$

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.

  $x^2+10x+25=0$

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

As $a=1,b=10$ and $c=25$

  $D={\left(10\right)}^2-4\times 1\times \left(25\right)$

Now,

  $D=100-100$

Hence,

  $D=0$

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

To determine

( $x^2+3x-12$)

Solve the equation.

Answer to Problem 41PPS

  $D=57$ and 2 real roots

Explanation of Solution

Given Information: $x^2+3x-12=0$

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.

  $x^2+3x-12=0$

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

As $a=1,b=3$ and $c=-12$

  $D={\left(3\right)}^2-4\times 1\times \left(-12\right)$

Now,

  $D=9+48$

Hence,

  $D=57$

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

To determine

(Table)

Solve the equation.

Answer to Problem 41PPS

  $\begin{array}{ccc}Trinomial& b^2-4ac& NumberofRoots\\ x^2-8x+16& 0& 1\\ 2x^2-11x+3& 97& 2\\ 3x^2+6x+9& -72& 0\\ x^2-2x+7& -24& 0\\ x^2+10x+25& 0& 1\\ x^2+3x-12& 57& 2\end{array}$

Explanation of Solution

Given Information: Equations

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