The zeros of the quadratic polynomial \[{x^2} + kx + k = 0,k > 0\]
A) Both cannot be positive
B) Both cannot be negative
C) Are always unequal
D) Are always equal
Answer
627.3k+ views
Hint: Use sridhar acharya's formula to judge the property of roots/zeros of the quadratic equation given. Take different examples and judge each case individually. The discriminant will play an important role.
Complete step by step answer:
We know that sridhar acharya's formula for quadratic equation is
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\] where discriminant D is given by \[D = {b^2} - 4ac\] in a general quadratic equation \[a{x^2} + bx + c = 0\]
In the given question \[a = 1\& b = c = k\] if we put the values in the discriminant formula we will get it as \[D = {k^2} - 4k\] . Now the discriminant must be greater than 0 or else the roots will not exist in the number line and if we put the whole thing in original sridharacharya formula we will get it as \[x = \dfrac{{ - k \pm \sqrt {{k^2} - 4k} }}{2}\] It is clear that if the \[\sqrt {{k^2} - 4k} > k\] then the root we will be getting one will be negative and one will be positive and if \[\sqrt {{k^2} - 4k} < k\] both the numbers will be negative so from this observation we can say that both the zeros cannot be positive but it can be negative Now let us check for option C and D
If we try to put \[k = 4\] in \[x = \dfrac{{ - k \pm \sqrt {{k^2} - 4k} }}{2}\] The square root part will vanish as \[{k^2} = {4^2} = 16\& 4k = 4 \times 4 = 16\] therefore \[D = 0\] So in this case the roots will be \[x = \dfrac{{ - 4 \pm 0}}{2} = - 2\] so the roots are equal but in any other case like if we take \[k = 5\& k = 6\] The roots will not be equal. Their options C and D are incorrect as all the roots are not always equal and at the sametime not always unequal.
This implies that the correct option is option A both roots cannot be positive. One of them has to be negative in any case.
Note: Always do this type of question by taking some examples as it helps to gain the accurate results. Don't try to do it theoretically or hypothetically. Always cross check your results by taking some numerical examples.
Complete step by step answer:
We know that sridhar acharya's formula for quadratic equation is
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\] where discriminant D is given by \[D = {b^2} - 4ac\] in a general quadratic equation \[a{x^2} + bx + c = 0\]
In the given question \[a = 1\& b = c = k\] if we put the values in the discriminant formula we will get it as \[D = {k^2} - 4k\] . Now the discriminant must be greater than 0 or else the roots will not exist in the number line and if we put the whole thing in original sridharacharya formula we will get it as \[x = \dfrac{{ - k \pm \sqrt {{k^2} - 4k} }}{2}\] It is clear that if the \[\sqrt {{k^2} - 4k} > k\] then the root we will be getting one will be negative and one will be positive and if \[\sqrt {{k^2} - 4k} < k\] both the numbers will be negative so from this observation we can say that both the zeros cannot be positive but it can be negative Now let us check for option C and D
If we try to put \[k = 4\] in \[x = \dfrac{{ - k \pm \sqrt {{k^2} - 4k} }}{2}\] The square root part will vanish as \[{k^2} = {4^2} = 16\& 4k = 4 \times 4 = 16\] therefore \[D = 0\] So in this case the roots will be \[x = \dfrac{{ - 4 \pm 0}}{2} = - 2\] so the roots are equal but in any other case like if we take \[k = 5\& k = 6\] The roots will not be equal. Their options C and D are incorrect as all the roots are not always equal and at the sametime not always unequal.
This implies that the correct option is option A both roots cannot be positive. One of them has to be negative in any case.
Note: Always do this type of question by taking some examples as it helps to gain the accurate results. Don't try to do it theoretically or hypothetically. Always cross check your results by taking some numerical examples.
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