Mixing
Solving quadratic inequality to get the final answer
Clash Royale CLAN TAG#URR8PPP
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Okay, so I'm self studying my A level mathematics and I don't have a tutor to ask from so I hope someone can help me here.
My question is:
Q: The equation of curve is $y=dfrac12x$ and the of a line $L$ is
$2 x + y= k$ where $k$ is a constant. Find the set of values of $k$ for which $L$ does not intersect the curve.
So here's my answer:
Ans:
•Solving simultaneously,
$$frac12x = kx - 2x^2$$
$$2x^2 - kx + 12 = 0$$
•Since $L$ doesn't intersect there are no real roots, therefore the discriminant is
$$b^2 - 4ac < 0$$
•substituting values
$$(-k)^2 - 4(2)(12) < 0$$
$$k^2 - 96 < 0$$
$$(k - sqrt96)(k + sqrt96) < 0$$
Now I don't know how to get the answer. I know to draw a graph, but for $x$ values, I don't understand how to use it for $k$ or neither do I know any other method.
(Can you keep the explanation simple in A level boundaries?)
inequality quadratics
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
asked Jul 29 at 13:11
Mushraf Altaf
62
add a comment |Â
up vote
1
down vote
favorite
Okay, so I'm self studying my A level mathematics and I don't have a tutor to ask from so I hope someone can help me here.
My question is:
Q: The equation of curve is $y=dfrac12x$ and the of a line $L$ is
$2 x + y= k$ where $k$ is a constant. Find the set of values of $k$ for which $L$ does not intersect the curve.
So here's my answer:
Ans:
•Solving simultaneously,
$$frac12x = kx - 2x^2$$
$$2x^2 - kx + 12 = 0$$
•Since $L$ doesn't intersect there are no real roots, therefore the discriminant is
$$b^2 - 4ac < 0$$
•substituting values
$$(-k)^2 - 4(2)(12) < 0$$
$$k^2 - 96 < 0$$
$$(k - sqrt96)(k + sqrt96) < 0$$
Now I don't know how to get the answer. I know to draw a graph, but for $x$ values, I don't understand how to use it for $k$ or neither do I know any other method.
(Can you keep the explanation simple in A level boundaries?)
inequality quadratics
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
asked Jul 29 at 13:11
Mushraf Altaf
62
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Okay, so I'm self studying my A level mathematics and I don't have a tutor to ask from so I hope someone can help me here.
My question is:
Q: The equation of curve is $y=dfrac12x$ and the of a line $L$ is
$2 x + y= k$ where $k$ is a constant. Find the set of values of $k$ for which $L$ does not intersect the curve.
So here's my answer:
Ans:
•Solving simultaneously,
$$frac12x = kx - 2x^2$$
$$2x^2 - kx + 12 = 0$$
•Since $L$ doesn't intersect there are no real roots, therefore the discriminant is
$$b^2 - 4ac < 0$$
•substituting values
$$(-k)^2 - 4(2)(12) < 0$$
$$k^2 - 96 < 0$$
$$(k - sqrt96)(k + sqrt96) < 0$$
Now I don't know how to get the answer. I know to draw a graph, but for $x$ values, I don't understand how to use it for $k$ or neither do I know any other method.
(Can you keep the explanation simple in A level boundaries?)
inequality quadratics
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
asked Jul 29 at 13:11
Mushraf Altaf
62
Okay, so I'm self studying my A level mathematics and I don't have a tutor to ask from so I hope someone can help me here.
My question is:
Q: The equation of curve is $y=dfrac12x$ and the of a line $L$ is
$2 x + y= k$ where $k$ is a constant. Find the set of values of $k$ for which $L$ does not intersect the curve.
So here's my answer:
Ans:
•Solving simultaneously,
$$frac12x = kx - 2x^2$$
$$2x^2 - kx + 12 = 0$$
•Since $L$ doesn't intersect there are no real roots, therefore the discriminant is
$$b^2 - 4ac < 0$$
•substituting values
$$(-k)^2 - 4(2)(12) < 0$$
$$k^2 - 96 < 0$$
$$(k - sqrt96)(k + sqrt96) < 0$$
Now I don't know how to get the answer. I know to draw a graph, but for $x$ values, I don't understand how to use it for $k$ or neither do I know any other method.
(Can you keep the explanation simple in A level boundaries?)
inequality quadratics
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
asked Jul 29 at 13:11
Mushraf Altaf
62
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
edited Jul 29 at 13:18
Karn Watcharasupat
3,7992426
3,7992426
asked Jul 29 at 13:11
Mushraf Altaf
62
asked Jul 29 at 13:11
Mushraf Altaf
62
asked Jul 29 at 13:11
Mushraf Altaf
62
62
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Your working so far is fine. Now you have
$$(k+sqrt96)(k-sqrt96)<0$$
To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-sqrt96<k<sqrt96$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your working so far is fine. Now you have
$$(k+sqrt96)(k-sqrt96)<0$$
To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-sqrt96<k<sqrt96$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
add a comment |Â
up vote
2
down vote
Your working so far is fine. Now you have
$$(k+sqrt96)(k-sqrt96)<0$$
To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-sqrt96<k<sqrt96$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your working so far is fine. Now you have
$$(k+sqrt96)(k-sqrt96)<0$$
To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-sqrt96<k<sqrt96$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
Your working so far is fine. Now you have
$$(k+sqrt96)(k-sqrt96)<0$$
To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-sqrt96<k<sqrt96$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
answered Jul 29 at 13:16
Karn Watcharasupat
3,7992426
3,7992426
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
add a comment |Â
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
Thank you for answering my question.
– Mushraf Altaf
Jul 29 at 13:30
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
@MushrafAltaf No worry! I just finished A Level a few months ago too :)
– Karn Watcharasupat
Jul 29 at 13:31
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 29 at 13:16