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Problems Involving Quadratic Inequalities

How do I solve Question 7 Exercise 3d?IMG_3136.jpg

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Original post by JudaicImposter
How do I solve Question 7 Exercise 3d?


The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.
(edited 5 years ago)
Original post by RDKGames
The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.

I still can't solve it
Original post by RDKGames
The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.

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


Would be simpler to just set the discriminant ≤0\leq 0.

Post your working if you get stuck.
Original post by RDKGames
The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.

How do I solve Question 9?
Original post by JudaicImposter
How do I solve Question 9?


x=2x=2 is a root. So plug that in and solve the equation for α\alpha first.
Original post by RDKGames
The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.


For Question 14b I've got -2x^2 -8x +10 for the first quadratic function, how do I find the second one?
(edited 5 years ago)
Original post by JudaicImposter
For Question 14b I've got -2x^2 +8x +10 for the first quadratic function, how do I find the second one?


I can't see Q14.
Original post by RDKGames
I can't see Q14.

I will scan the page and upload it...
Original post by RDKGames
I can't see Q14.

Another scanpg49_new.jpg
Original post by JudaicImposter
For Question 14b I've got -2x^2 +8x +10 for the first quadratic function, how do I find the second one?


That's not right, you might want to check the sign of that 8x8x.

OK so we can begin by saying we are seeking a quadratic function ax2+bx+cax^2 + bx + c such that:

a+b+c=0a+b+c = 0 (i.e. x=1 is a root)

0+0+c=100+0+c=10 (i.e. takes the value of 10 when x=0) hence c=10c=10, and the first equation implies a+b=−10a+b = -10.

a(−b2a)2+b(−b2a)+c=18  ⟹  b24a−b22a=8a(\frac{-b}{2a})^2 + b (-\frac{b}{2a}) + c = 18 \implies \frac{b^2}{4a} - \frac{b^2}{2a} = 8 (i.e. extremal value of 18)

a<0a<0 (we make this extremal value to be a maximum)

So, from the third equation we get that −b2=32a- b^2 = 32a which we can sub into a+b=−10  ⟹  32a+32b=−320  ⟹  −b2+32b=−320a+b = -10 \implies 32a+32b = -320 \implies -b^2 +32b = -320.

This clearly gives two values of bb, hence you can anticipate two quadratics, and indeed both values give us -ve aa as required since a=−b232<0a=-\frac{b^2}{32}<0.
(edited 5 years ago)
Original post by RDKGames
That's not right, you might want to check the sign of that 8x8x.

OK so we can begin by saying we are seeking a quadratic function ax2+bx+cax^2 + bx + c such that:

a+b+c=0a+b+c = 0 (i.e. x=1 is a root)

0+0+c=100+0+c=10 (i.e. takes the value of 10 when x=0) hence c=10c=10, and the first equation implies a+b=−10a+b = -10.

a(−b2a)2+b(−b2a)+c=18  ⟹  b24a−b22a=8a(\frac{-b}{2a})^2 + b (-\frac{b}{2a}) + c = 18 \implies \frac{b^2}{4a} - \frac{b^2}{2a} = 8 (i.e. extremal value of 18)

a<0a<0 (we make this extremal value to be a maximum)

So, from the third equation we get that −b2=32a- b^2 = 32a which we can sub into a+b=−10  ⟹  32a+32b=−320  ⟹  −b2+32b=−320a+b = -10 \implies 32a+32b = -320 \implies -b^2 +32b = -320.

This clearly gives two values of bb, hence you can anticipate two quadratics, and indeed both values give us -ve aa as required since a=−b232<0a=-\frac{b^2}{32}<0.


I've now corrected the sign of 8x
Original post by RDKGames
The leading coefficient must be strictly +ve first of all, so here it is 2+k2+k hence 2+k>02+k > 0.

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 kk.


I'm now stuck on Question 17b. Please provide help.
Original post by JudaicImposter
I'm now stuck on Question 17b. Please provide help.


You are seeking kk such that k(x+2)2−(x−1)(x−2)≤12.5k(x+2)^2-(x-1)(x-2) \leq 12.5

Hence k(x+2)2−(x−1)(x−2)−12.5≤0k(x+2)^2-(x-1)(x-2) - 12.5 \leq 0

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 kk using that.
Also some simple analysis reveals that the coeff of x2x^2 is k−1k-1 and you want this to be <0 hence k<1k<1 is necessary.
(edited 5 years ago)
Original post by RDKGames
You are seeking kk such that k(x+2)2−(x−1)(x−2)≤12.5k(x+2)^2-(x-1)(x-2) \leq 12.5

Hence k(x+2)2−(x−1)(x−2)−12.5≤0k(x+2)^2-(x-1)(x-2) - 12.5 \leq 0

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 kk using that.
Also some simple analysis reveals that the coeff of x2x^2 is k−1k-1 and you want this to be <0 hence k<1k<1 is necessary.


I've only got k is less or equal to 0 as the answer. Is that correct? I expanded the two quadratics, determined that the graph has a maximum at x equal to 0 , the inequality is true for all real x so i substituted an arbitrary small value for x, specifically the value of 2, got an inequality involving only a multiple of k on the left hand side and less than or equal to 0 on the right. It doesn't look like I've completely finished the answer.
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?
(edited 5 years ago)
Original post by RDKGames
You are seeking kk such that k(x+2)2−(x−1)(x−2)≤12.5k(x+2)^2-(x-1)(x-2) \leq 12.5

Hence k(x+2)2−(x−1)(x−2)−12.5≤0k(x+2)^2-(x-1)(x-2) - 12.5 \leq 0

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 kk using that.
Also some simple analysis reveals that the coeff of x2x^2 is k−1k-1 and you want this to be <0 hence k<1k<1 is necessary.


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


No. Max value of the function with that value of kk is 44 which exceeds the 12.5 limit.
Original post by RDKGames
No. Max value of the function with that value of kk is 44 which exceeds the 12.5 limit.


I've given up. Please post the solution for the whole question including for the two graphs at the end. Thanks.
(edited 5 years ago)
Original post by RDKGames
You are seeking kk such that k(x+2)2−(x−1)(x−2)≤12.5k(x+2)^2-(x-1)(x-2) \leq 12.5

Hence k(x+2)2−(x−1)(x−2)−12.5≤0k(x+2)^2-(x-1)(x-2) - 12.5 \leq 0

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 kk using that.
Also some simple analysis reveals that the coeff of x2x^2 is k−1k-1 and you want this to be <0 hence k<1k<1 is necessary.


Couldn't an alternative be that the left hand side of the second equation means there are two real roots and the fact that it is an inequality means that the equation dips below the x axis for a given range of x values?

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