Consider a quadratic equationx2+2x=k+∫01|t+k|dtThen choose the correct option(s),A) Roots are RealB) Roots are ImaginaryC) Roots are DistinctD) Roots are complex number

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Mikaela Winters · Accepted Answer

Step 1The discriminant b2−4ac is actually1) d=4+4(k+∫01|t+k|dt)Regarding calculating k+∫01|t+k|dt, there are 3 cases to consider:k&amp;lt;−1Since t+k&amp;lt;0 for t from 0 to 1, this givesk+∫01|t+k|dt=k+∫01(−t−k)dt2) =k+(−t22−kt)∣01=k+(−12−k)=−12In (1), this gives d=2Step 2−1≤k&amp;lt;0This requires splitting the integration into 2 parts, the initial one for t from 0 to -k wheret+k≤0⇒|t+k|=−t−kand a second part for t from -k to 1 where t+k≥0⇒|t+k|=t+kgivingk+∫01|t+k|dt=k+∫0−k(−t−k)dt+∫−k1(t+k)dt=k(−t22−kt)|{0}−k+(t22+kt)|−k13) =k−k22+k2+12+k−k22+k2=2k+k2+12{2}+2k+1−1+12=(k+1)2−12or k≥−1, the value above is an increasing function, going from −12 to 12. Thus, d in (1) goes from 2 to 6.Step 3k≥0This obviously gives d&amp;gt;0, but doing the calculations results ink+∫01|t+k|dt=k+∫01(t+k)dt4) =k+(t22+kt)∣01=k+(12+k)=2k+12Thus, d in (1) is 6 at k=0, and increases without bound as k goes up.This shows all 3 cases gives d&amp;gt;0, so the roots are real and distinct. Since the roots are also always complex numbers, this means options (A), (C) and (D) are true, with only (B) being false since there are no imaginary roots.

microsgopx6z7 · Accepted Answer

Step 1As pointed out in comments and in John Omielan&#039;s answer, to show that there are always two roots, it suffices to show that 1+k+∫01|t+k|dt≥0for ever k∈R. From triangle inequality (or reverse triangle inequality) we get that |t+k|≥|k|−|t| Since we are looking only at t between 0 and 1, we can actually use|t+k|≥|k|−t to get1+k+∫01|t+k|dt≥1+k+∫01(|k|−t)dt=1+k+∫01|k|dt−∫01t dt=1+k+|k|−12=12+k+|k|≥12(We have used x+|x|≥0.)I wouldn&#039;t be trying to find some kind of proof which does not need to be split in the cases if I hadn&#039;t seen in the other answer that this always has two distinct real roots. So the credit should go there.As a side note, I will also mention that there is this older questions: Comment upon the nature of roots of the quadratic equation x2+2x=k+∫01|t+k|dt depending on the value of k∈R However, it is now deleted, so it&#039;s visible only to 10k+ users. One interesting thing I see there is that ∫|x|dx=x×|x|2 is used there to evaluate the integral.

Consider a quadratic equation \(\displaystyle{x}^{{{2}}}+{2}{x}={k}+{\int_{{{0}}}^{{{1}}}}{\left|{t}+{k}\right|}{\left.{d}{t}\right.}\) Then choose the correct

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