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Bismuth83
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Given the equation x^2kx+20=0 , where k is a constant, which of the 16 Oct 2024, 15:24
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k: 9 x: 4
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71% (01:55) correct 29% (02:04) wrong based on 1419 sessions

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Given the equation \(x^2−kx+20=0\), where \(k\) is a constant, which of the following could be corresponding values of integers \(k\) and \(x\)?­
k x
-20
-9
-2
0
4
9
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Given the equation x^2kx+20=0 , where k is a constant, which of the 23 Oct 2024, 09:40
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Bismuth83
Given the equation \(x^2−kx+20=0\), where \(k\) is a constant, which of the following could be corresponding values of integers \(k\) and \(x\)?­
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A very quick method and a method not getting into quadratic equations will be..

\(x^2−kx+20=0......x(x-k)=-20=-1×20=-2×10=-4×5\)

The products that are in options..
1. -20 : x(x-k) = 1×-20 ... This would require k to be -21....-20×(-20-(-21))
21 is not in the options.

2. -2 : x(x-k) = -2×10... This would require k to be -12....-2×(-2-(-12))
10 is not in the options.

3. 4 : x(x-k) = 4×-5 ... This would require k to be 9....4×(4-9).... We have both 4 and 9 in the options.

Thus x=4 & k=9
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Given the equation x^2kx+20=0 , where k is a constant, which of the 17 Oct 2024, 06:07
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Bismuth83
Given the equation \(x^2−kx+20=0\), where \(k\) is a constant, which of the following could be corresponding values of integers \(k\) and \(x\)?­
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Solution: x^2 -kx+20 = 0

We have to factorize and find two solutions for x. Here you go:


(x-4) (x-5) = 0

We know x could be either 4 or 5. Now, as per the value given to us the value for x is 4.

We can put this value into the equation x^2 -kx+20 = 0 and have the value for constant K is 9.

Hence, the answer is x= 4 and K = 9
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Given the equation x^2kx+20=0 , where k is a constant, which of the 17 Oct 2024, 06:25
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Use quadratic qns to solve qn
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Given the equation x^2kx+20=0 , where k is a constant, which of the 17 Oct 2024, 10:46
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Given the equation \(x^2−kx+20=0\), where \(k\) is a constant, which of the following could be corresponding values of integers \(k\) and \(x\)?­
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1. We are asked to find what could be k and x in this equation.

2. The first idea is to use the quadratic formula: \(x_{1,2} = \frac{k \pm \sqrt{k^2 - 80}}{2}\).

3. Since the square root takes only nonnegative numbers, \(k^2 - 80 \geq 0 \rightarrow k^2 \geq 80 \rightarrow |k| \geq \sqrt{80}\). Looking at the answer choices, k is whole, so \(|k| \geq \sqrt{81} = 9\). The only possible answer choices are k = -20 and \(\pm\)9.

4. It can also be noticed that \(\sqrt{k^2 - 80}\) must be whole for x to be whole too. \(\pm\)9 work, however, if k = -20 then \(\sqrt{k^2 - 80} = \sqrt{400 - 80} = \sqrt{320} = 8\sqrt{5}\), which isn't whole. So, only \(\pm\)9 could work.

5. Plugging k = -9 gives the roots -5 and -4, while k = 9 gives 5 and 4. Only 4 is a possible x.

6. Our answer will be k = 9 and x = 4.

__________________________________

Another way of excluding all options except k = -20 and \(\pm\)9 is by drawing out the two graphs \(x^2 + 20\) and \(kx\). They should be equal and it's easy to realize that k has to be large (absolute value) for the two graphs to intersect.
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Given the equation x^2kx+20=0 , where k is a constant, which of the 17 Oct 2024, 10:59
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naveeng15
Use quadratic qns to solve qn
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Yes, I knew the quadratic Equations formula to solve this ques but the only problem is that it consumes a lot of time to solve the ques and get to the solution. My apologies. Next, time I will take care of these steps to provide a solution. Thanks for the advice though.
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Given the equation x^2kx+20=0 , where k is a constant, which of the 22 Nov 2025, 03:09
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