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The Quadratic Formula: Mastery Test

Select the correct answer from each drop-down menu.

Determine the number of real solutions for each of the given equations.

\begin{tabular}{|c|c|}
\hline Equation & Number of Solutions \\
\hline [tex]$y=-3x^2+x+12$[/tex] & \\
\hline [tex]$y=2x^2-6x+5$[/tex] & \\
\hline [tex]$y=x^2+7x-11$[/tex] & \\
\hline [tex]$y=-x^2-8x-16$[/tex] & \\
\hline
\end{tabular}

Sagot :

GinnyAnsweridn
To determine the number of real solutions for each of the given quadratic equations, we analyze the discriminant (Δ), which is given by the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients from the quadratic equation in the standard form [tex]\( y = ax^2 + bx + c \)[/tex].

### Equation 1: [tex]\( y = -3x^2 + x + 12 \)[/tex]
For this equation:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = 12 \)[/tex]

Calculate the discriminant:
[tex]\[ \Delta = 1^2 - 4(-3)(12) = 1 + 144 = 145 \][/tex]
Since [tex]\( \Delta > 0 \)[/tex], there are 2 real solutions.

### Equation 2: [tex]\( y = 2x^2 - 6x + 5 \)[/tex]
For this equation:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -6 \)[/tex]
- [tex]\( c = 5 \)[/tex]

Calculate the discriminant:
[tex]\[ \Delta = (-6)^2 - 4(2)(5) = 36 - 40 = -4 \][/tex]
Since [tex]\( \Delta < 0 \)[/tex], there are 0 real solutions.

### Equation 3: [tex]\( y = x^2 + 7x - 11 \)[/tex]
For this equation:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 7 \)[/tex]
- [tex]\( c = -11 \)[/tex]

Calculate the discriminant:
[tex]\[ \Delta = 7^2 - 4(1)(-11) = 49 + 44 = 93 \][/tex]
Since [tex]\( \Delta > 0 \)[/tex], there are 2 real solutions.

### Equation 4: [tex]\( y = -x^2 - 8x - 16 \)[/tex]
For this equation:
- [tex]\( a = -1 \)[/tex]
- [tex]\( b = -8 \)[/tex]
- [tex]\( c = -16 \)[/tex]

Calculate the discriminant:
[tex]\[ \Delta = (-8)^2 - 4(-1)(-16) = 64 - 64 = 0 \][/tex]
Since [tex]\( \Delta = 0 \)[/tex], there is 1 real solution.

### Summary
\begin{tabular}{|c|c|}
\hline Equation & Number of Solutions \\
\hline[tex]$y=-3 x^2+x+12$[/tex] & 2 \\
\hline[tex]$y=2 x^2-6 x+5$[/tex] & 0 \\
\hline[tex]$y=x^2+7 x-11$[/tex] & 2 \\
\hline[tex]$y=-x^2-8 x-16$[/tex] & 1 \\
\hline
\end{tabular}

Therefore, the number of real solutions for each of the given quadratic equations is:
- [tex]\( y = -3 x^2 + x + 12 \)[/tex]: 2 solutions
- [tex]\( y = 2 x^2 - 6 x + 5 \)[/tex]: 0 solutions
- [tex]\( y = x^2 + 7 x - 11 \)[/tex]: 2 solutions
- [tex]\( y = -x^2 - 8 x - 16 \)[/tex]: 1 solution
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