Easier way to find amount of solutions between a line and quadratic?
Is there a better way of finding the amount of solutions between $y=(x-7)(3x+4)$ and $x=3y-1$ that doesn't involve calculus?
I know it is two but that's due to substitution of one equation into the other and finding the discriminant of the resulting quadratic, but it's kinda tedious and this is suppose to be a question that can be done within a minute.
Can I just say the vertex of the upward-opening parabola is below the y-intercept of the line and this would result in 2 intersections?
quadratics
asked 2 hours ago
Sat
325
add a comment |
Is there a better way of finding the amount of solutions between $y=(x-7)(3x+4)$ and $x=3y-1$ that doesn't involve calculus?
I know it is two but that's due to substitution of one equation into the other and finding the discriminant of the resulting quadratic, but it's kinda tedious and this is suppose to be a question that can be done within a minute.
Can I just say the vertex of the upward-opening parabola is below the y-intercept of the line and this would result in 2 intersections?
quadratics
asked 2 hours ago
Sat
325
1
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
1
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
2
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
2
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
2
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago
add a comment |
Is there a better way of finding the amount of solutions between $y=(x-7)(3x+4)$ and $x=3y-1$ that doesn't involve calculus?
I know it is two but that's due to substitution of one equation into the other and finding the discriminant of the resulting quadratic, but it's kinda tedious and this is suppose to be a question that can be done within a minute.
Can I just say the vertex of the upward-opening parabola is below the y-intercept of the line and this would result in 2 intersections?
quadratics
asked 2 hours ago
Sat
325
Is there a better way of finding the amount of solutions between $y=(x-7)(3x+4)$ and $x=3y-1$ that doesn't involve calculus?
I know it is two but that's due to substitution of one equation into the other and finding the discriminant of the resulting quadratic, but it's kinda tedious and this is suppose to be a question that can be done within a minute.
Can I just say the vertex of the upward-opening parabola is below the y-intercept of the line and this would result in 2 intersections?
quadratics
quadratics
asked 2 hours ago
Sat
325
asked 2 hours ago
Sat
325
asked 2 hours ago
Sat
325
asked 2 hours ago
Sat
325
asked 2 hours ago
Sat
325
325
1
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
1
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
2
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
2
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
2
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago
add a comment |
1
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
1
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
2
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
2
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
2
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago
1
1
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
1
1
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
2
2
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
2
2
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
2
2
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago
add a comment |
2 Answers
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A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).
answered 1 hour ago
Cactus
558
add a comment |
It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).
In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = frac{1}{3} x + frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - frac{52}{3} x - frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.
Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.
answered 1 hour ago
Travis
59.7k767146
add a comment |
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2 Answers
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A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).
answered 1 hour ago
Cactus
558
add a comment |
A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).
answered 1 hour ago
Cactus
558
add a comment |
A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).
answered 1 hour ago
Cactus
558
A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).
answered 1 hour ago
Cactus
558
answered 1 hour ago
Cactus
558
answered 1 hour ago
Cactus
558
answered 1 hour ago
Cactus
558
558
add a comment |
add a comment |
It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).
In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = frac{1}{3} x + frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - frac{52}{3} x - frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.
Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.
answered 1 hour ago
Travis
59.7k767146
add a comment |
It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).
In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = frac{1}{3} x + frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - frac{52}{3} x - frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.
Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.
answered 1 hour ago
Travis
59.7k767146
add a comment |
It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).
In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = frac{1}{3} x + frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - frac{52}{3} x - frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.
Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.
answered 1 hour ago
Travis
59.7k767146
It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).
In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = frac{1}{3} x + frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - frac{52}{3} x - frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.
Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.
answered 1 hour ago
Travis
59.7k767146
answered 1 hour ago
Travis
59.7k767146
answered 1 hour ago
Travis
59.7k767146
answered 1 hour ago
Travis
59.7k767146
59.7k767146
add a comment |
add a comment |
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1
I guess technically you would need to justify why an upward opening parabola whose vertex (centered on the $y$-axis) lies below the $y$-intercept of a line is always guaranteed to intersect said line twice, but yes, that argument is good. Is there a particular reason you want to avoid solving the quadratic formula? You should be able to compute such a thing in under a minute quite easily.
– ItsJustASeriesBro
2 hours ago
1
Yes, this works too (though I wouldn't recommend answering an exam question like that).
– John Doe
2 hours ago
2
As a final note, your parabola's vertex is not centered on the $y$-axis, and so you won't even be able to use the "theorem" you want to in order to take this shortcut.
– ItsJustASeriesBro
2 hours ago
2
@ItsJustASeriesBro Technically, all you need to do is find any value of $x$ for which the parabola is below the line, and you're done. No need for it to be the vertex. But I know what you mean, this would be a more general theorem.
– John Doe
2 hours ago
2
@JohnDoe Of course, but the OP specifically wanted to use the vertex. And proving this theorem seems to be more complicated than using the quadratic formula to me in any event.
– ItsJustASeriesBro
2 hours ago