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Sagot :
To find the points of intersection between the two equations, we first set them equal to each other and solve for [tex]\( x \)[/tex]:
[tex]\[ y = 3x + 17 \][/tex]
[tex]\[ y = -2x^2 - 12x - 16 \][/tex]
Set the right-hand sides equal to each other:
[tex]\[ 3x + 17 = -2x^2 - 12x - 16 \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ 2x^2 + 12x + 3x + 16 + 17 = 0 \][/tex]
[tex]\[ 2x^2 + 15x + 33 = 0 \][/tex]
Next, we solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 15 \)[/tex], and [tex]\( c = 33 \)[/tex].
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 15^2 - 4(2)(33) \][/tex]
[tex]\[ \Delta = 225 - 264 \][/tex]
[tex]\[ \Delta = -39 \][/tex]
Since the discriminant is negative, there are no real solutions to the equation. This means that the curves do not intersect at any real points.
Therefore, the answer is:
[tex]\[ (x, y) = \text{No solution} \][/tex]
[tex]\[ y = 3x + 17 \][/tex]
[tex]\[ y = -2x^2 - 12x - 16 \][/tex]
Set the right-hand sides equal to each other:
[tex]\[ 3x + 17 = -2x^2 - 12x - 16 \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ 2x^2 + 12x + 3x + 16 + 17 = 0 \][/tex]
[tex]\[ 2x^2 + 15x + 33 = 0 \][/tex]
Next, we solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 15 \)[/tex], and [tex]\( c = 33 \)[/tex].
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 15^2 - 4(2)(33) \][/tex]
[tex]\[ \Delta = 225 - 264 \][/tex]
[tex]\[ \Delta = -39 \][/tex]
Since the discriminant is negative, there are no real solutions to the equation. This means that the curves do not intersect at any real points.
Therefore, the answer is:
[tex]\[ (x, y) = \text{No solution} \][/tex]
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