Suppose the equivalent quadratic equation is written x2−18x+c=−80+c

Question

. What value of c makes the equation a perfect square trinomial?

Answer 1

To make the equation \( x^2 - 18x + c = -80 + c \) into a perfect square trinomial, we first need to simplify it.

Let’s rewrite the equation as:

\[ x^2 - 18x + c + 80 - c = 0 \]

This simplifies to:

\[ x^2 - 18x + 80 = 0 \]

Now, we want to express \( x^2 - 18x + 80 \) as a perfect square. A quadratic in the form \( x^2 - bx \) can be turned into a perfect square by finding the needed constant.

To achieve the form \( (x - k)^2 \), where \( k \) is half of the coefficient of \( x \):

1. The coefficient of \( x \) is \( -18 \), so half of that is \(-9\).
2. To create a perfect square, we need to add \( (-9)^2 \) or \( 81 \).

Thus, we adjust \( c \) in our original equation to ensure that:

\[ x^2 - 18x + c = (x - 9)^2 \]

Here we set:

\[ c = 81 - 80 \]

So we need to balance this equation correctly:

If we require our trinomial to be a perfect square, it should actually be:

\[ x^2 - 18x + c - 80 = 0 \]

Now substituting \( c = 81 \):

\[ x^2 - 18x + 81 - 80 = 0 \ x^2 - 18x + 1 = (x - 9)^2 - 80 = 0 \ \]

This shows that for \( c = 81 \), the equation becomes \( (x - 9)^2 = 0\), confirming that it is a perfect square trinomial.

Thus, the value of \( c \) that makes the equation a perfect square trinomial is:

\[ \boxed{81} \]

Answer 2

5. Which of the following quadratic equations is not solvable by grouping?(1 point)
Responses

2x2−2x−10=0
2 x squared minus 2 x minus 10 is equal to 0

2x2+14x+12=0
2 x squared plus 14 x plus 12 is equal to 0

x2−2x+1=0
x squared minus 2 x plus 1 is equal to 0

x2−12x+35=0

Answer 3

To determine which quadratic equation is not solvable by grouping, we should look at the factorization of each option. Grouping typically involves rearranging terms and factoring them in pairs.

Let’s evaluate each of the given equations:

1. \(2x^2 - 2x - 10 = 0\)

   • We can factor out a 2: \[ 2(x^2 - x - 5) = 0 \]
   • The quadratic \(x^2 - x - 5\) does not factor neatly, nor can it be simplified via grouping.

2. \(2x^2 + 14x + 12 = 0\)

   • We can factor out a 2: \[ 2(x^2 + 7x + 6) = 0 \]
   • \(x^2 + 7x + 6\) factors to \((x + 6)(x + 1)\).

3. \(x^2 - 2x + 1 = 0\)

   • This is a perfect square: \[ (x - 1)^2 = 0 \]

4. \(x^2 - 12x + 35 = 0\)

   • We can factor this quadratic: \[ (x - 5)(x - 7) = 0 \]

Analysis:

• The first equation \(2x^2 - 2x - 10 = 0\) does not factor neatly and is not easily solvable by grouping since \(x^2 - x - 5\) cannot be factored into rational roots.
• The second, third, and fourth equations can all be factored (to some extent) and are solvable.

Conclusion:

The quadratic equation that is not solvable by grouping is:

\[ \boxed{2x^2 - 2x - 10 = 0} \]