Given the equation x^2kx+20=0 , where k is a constant, which of the

Question

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 ?

chetan2u · Accepted Answer

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

poojaarora1818 · Answer

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

naveeng15 · Answer

Use quadratic qns to solve qn

Bismuth83 · Answer

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.

poojaarora1818 · Answer

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

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