# Problems Involving Quadratic Inequalities

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#2

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How do I solve Question 7 Exercise 3d?

**JudaicImposter**)How do I solve Question 7 Exercise 3d?

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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(Original post by

The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

**RDKGames**)The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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**RDKGames**)

The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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#5

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The function has a minimum value of -36/(2+k) + (2k+1) now what do i do?

**JudaicImposter**)The function has a minimum value of -36/(2+k) + (2k+1) now what do i do?

Post your working if you get stuck.

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**RDKGames**)

The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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#7

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How do I solve Question 9?

**JudaicImposter**)How do I solve Question 9?

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**RDKGames**)

The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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#9

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For Question 14b I've got -2x^2 +8x +10 for the first quadratic function, how do I find the second one?

**JudaicImposter**)For Question 14b I've got -2x^2 +8x +10 for the first quadratic function, how do I find the second one?

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(Original post by

I can't see Q14.

**RDKGames**)I can't see Q14.

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#12

**JudaicImposter**)

For Question 14b I've got -2x^2 +8x +10 for the first quadratic function, how do I find the second one?

OK so we can begin by saying we are seeking a quadratic function such that:

(i.e. x=1 is a root)

(i.e. takes the value of 10 when x=0) hence , and the first equation implies .

(i.e. extremal value of 18)

(we make this extremal value to be a maximum)

So, from the third equation we get that which we can sub into .

This clearly gives two values of , hence you can anticipate two quadratics, and indeed both values give us -ve as required since .

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(Original post by

That's not right, you might want to check the sign of that .

OK so we can begin by saying we are seeking a quadratic function such that:

(i.e. x=1 is a root)

(i.e. takes the value of 10 when x=0) hence , and the first equation implies .

(i.e. extremal value of 18)

(we make this extremal value to be a maximum)

So, from the third equation we get that which we can sub into .

This clearly gives two values of , hence you can anticipate two quadratics, and indeed both values give us -ve as required since .

**RDKGames**)That's not right, you might want to check the sign of that .

OK so we can begin by saying we are seeking a quadratic function such that:

(i.e. x=1 is a root)

(i.e. takes the value of 10 when x=0) hence , and the first equation implies .

(i.e. extremal value of 18)

(we make this extremal value to be a maximum)

So, from the third equation we get that which we can sub into .

This clearly gives two values of , hence you can anticipate two quadratics, and indeed both values give us -ve as required since .

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**RDKGames**)

The leading coefficient must be strictly +ve first of all, so here it is hence .

And so now that our quadratic is of U shape, we also require that this quadratic either has one real root, or none. Or in other words, it has no two distinct roots (otherwise there is a region between the roots where the quadratic dips into -ve values of y which is not what we want). This is to make sure that our quadratic is entirely above (or touching) the x-axis. So use this to determine the stricter condition for .

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#15

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I'm now stuck on Question 17b. Please provide help.

**JudaicImposter**)I'm now stuck on Question 17b. Please provide help.

Hence

If the LHS is less than or equal to zero, then that means it has either a single root, or no roots. So find the values of using that.

Also some simple analysis reveals that the coeff of is and you want this to be <0 hence is necessary.

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(Original post by

You are seeking such that

Hence

If the LHS is less than or equal to zero, then that means it has either a single root, or no roots. So find the values of using that.

Also some simple analysis reveals that the coeff of is and you want this to be <0 hence is necessary.

**RDKGames**)You are seeking such that

Hence

If the LHS is less than or equal to zero, then that means it has either a single root, or no roots. So find the values of using that.

Also some simple analysis reveals that the coeff of is and you want this to be <0 hence is necessary.

Edit: I think I've got it now; I had forgotten the -12.5 on the left hand side

Second Edit: The final answer is now k is less than or equal to (25/32). Is that correct?

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**RDKGames**)

You are seeking such that

Hence

If the LHS is less than or equal to zero, then that means it has either a single root, or no roots. So find the values of using that.

Also some simple analysis reveals that the coeff of is and you want this to be <0 hence is necessary.

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#18

(Original post by

Is k is less than or equal to (25/32) correct?

**JudaicImposter**)Is k is less than or equal to (25/32) correct?

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(Original post by

No. Max value of the function with that value of is 44 which exceeds the 12.5 limit.

**RDKGames**)No. Max value of the function with that value of is 44 which exceeds the 12.5 limit.

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**RDKGames**)

You are seeking such that

Hence

If the LHS is less than or equal to zero, then that means it has either a single root, or no roots. So find the values of using that.

Also some simple analysis reveals that the coeff of is and you want this to be <0 hence is necessary.

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