Question
Algebra 2A Semester Exam
27 of 4027 of 40 Items
Question
Solve the system of equations.
y=2x2−9
y=x2−10x−25
Enter the solution with the least x-value first.
(1 point)
The solutions are
,
, and
,
.
Answers
GPT-4o mini
To solve the system of equations given by:
1. \( y = 2x^2 - 9 \)
2. \( y = x^2 - 10x - 25 \)
we can set the two equations equal to each other since they both equal \( y \):
\[
2x^2 - 9 = x^2 - 10x - 25
\]
Now, let's move all terms to one side of the equation:
\[
2x^2 - x^2 + 10x - 25 + 9 = 0
\]
\[
x^2 + 10x - 16 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 10 \), and \( c = -16 \).
Calculating the discriminant:
\[
b^2 - 4ac = 10^2 - 4(1)(-16) = 100 + 64 = 164
\]
Now we can plug in the values into the quadratic formula:
\[
x = \frac{-10 \pm \sqrt{164}}{2(1)} = \frac{-10 \pm \sqrt{164}}{2}
\]
Simplifying \( \sqrt{164} \):
\[
\sqrt{164} = \sqrt{4 \cdot 41} = 2\sqrt{41}
\]
So, we have:
\[
x = \frac{-10 \pm 2\sqrt{41}}{2} = -5 \pm \sqrt{41}
\]
This gives us the two solutions for \( x \):
\[
x_1 = -5 + \sqrt{41}
\]
\[
x_2 = -5 - \sqrt{41}
\]
Now we calculate the corresponding \( y \) values by substituting back into one of the original equations, for example \( y = 2x^2 - 9 \).
For \( x_1 = -5 + \sqrt{41} \):
\[
y_1 = 2(-5 + \sqrt{41})^2 - 9
\]
Calculating \( (-5 + \sqrt{41})^2 \):
\[
(-5 + \sqrt{41})^2 = 25 - 10\sqrt{41} + 41 = 66 - 10\sqrt{41}
\]
Thus,
\[
y_1 = 2(66 - 10\sqrt{41}) - 9 = 132 - 20\sqrt{41} - 9 = 123 - 20\sqrt{41}
\]
For \( x_2 = -5 - \sqrt{41} \):
\[
y_2 = 2(-5 - \sqrt{41})^2 - 9
\]
Calculating \( (-5 - \sqrt{41})^2 \):
\[
(-5 - \sqrt{41})^2 = 25 + 10\sqrt{41} + 41 = 66 + 10\sqrt{41}
\]
Thus,
\[
y_2 = 2(66 + 10\sqrt{41}) - 9 = 132 + 20\sqrt{41} - 9 = 123 + 20\sqrt{41}
\]
Now we have our solutions:
1. \( \left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \)
2. \( \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right) \)
Therefore, the solutions in order of least \( x \)-value first:
\[
\left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \quad \text{and} \quad \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right)
\]
1. \( y = 2x^2 - 9 \)
2. \( y = x^2 - 10x - 25 \)
we can set the two equations equal to each other since they both equal \( y \):
\[
2x^2 - 9 = x^2 - 10x - 25
\]
Now, let's move all terms to one side of the equation:
\[
2x^2 - x^2 + 10x - 25 + 9 = 0
\]
\[
x^2 + 10x - 16 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 10 \), and \( c = -16 \).
Calculating the discriminant:
\[
b^2 - 4ac = 10^2 - 4(1)(-16) = 100 + 64 = 164
\]
Now we can plug in the values into the quadratic formula:
\[
x = \frac{-10 \pm \sqrt{164}}{2(1)} = \frac{-10 \pm \sqrt{164}}{2}
\]
Simplifying \( \sqrt{164} \):
\[
\sqrt{164} = \sqrt{4 \cdot 41} = 2\sqrt{41}
\]
So, we have:
\[
x = \frac{-10 \pm 2\sqrt{41}}{2} = -5 \pm \sqrt{41}
\]
This gives us the two solutions for \( x \):
\[
x_1 = -5 + \sqrt{41}
\]
\[
x_2 = -5 - \sqrt{41}
\]
Now we calculate the corresponding \( y \) values by substituting back into one of the original equations, for example \( y = 2x^2 - 9 \).
For \( x_1 = -5 + \sqrt{41} \):
\[
y_1 = 2(-5 + \sqrt{41})^2 - 9
\]
Calculating \( (-5 + \sqrt{41})^2 \):
\[
(-5 + \sqrt{41})^2 = 25 - 10\sqrt{41} + 41 = 66 - 10\sqrt{41}
\]
Thus,
\[
y_1 = 2(66 - 10\sqrt{41}) - 9 = 132 - 20\sqrt{41} - 9 = 123 - 20\sqrt{41}
\]
For \( x_2 = -5 - \sqrt{41} \):
\[
y_2 = 2(-5 - \sqrt{41})^2 - 9
\]
Calculating \( (-5 - \sqrt{41})^2 \):
\[
(-5 - \sqrt{41})^2 = 25 + 10\sqrt{41} + 41 = 66 + 10\sqrt{41}
\]
Thus,
\[
y_2 = 2(66 + 10\sqrt{41}) - 9 = 132 + 20\sqrt{41} - 9 = 123 + 20\sqrt{41}
\]
Now we have our solutions:
1. \( \left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \)
2. \( \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right) \)
Therefore, the solutions in order of least \( x \)-value first:
\[
\left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \quad \text{and} \quad \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right)
\]