Let x1 and x2 be the real root of the equation x2−kx+(k2+7k+15)=0, if the maximum value of (x12+x22)=d18x, then find the value of x?

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Let x1 and x2 be the real root of the equation x2−kx+(k2+7k+15)=0, if the maximum value of (x12+x22)=d18x, then find the value of x?

exinnaemekswr1k · Accepted Answer

First, it seems like you&#039;re using the symbol x to denote 2 different things (the variable in the quadratic equation, and the number we need in 18x). Avoid doing that because it causes confusion.Next, the problem with your solution is that you have a condition on k. The discriminant D=−3k2−28k−60 has to be non-negative for the quadratic to have real roots. Solving D≥0 gives k∈[−6,−103]. So, you have to find the maximum of f(k)=−k2−14k−30 on [−6,−103]. Instead, you found the maximum over R which is obtained when k=−7, and that&#039;s not in the interval. That&#039;s why you got a wrong answer. You&#039;ll find that f is decreasing in [−6,−103], so the maximum is f(−6)=18, and the answer follows.

Let \(\displaystyle{x}_{{{1}}}\ \text{and}\ {x}_{{{2}}}\) be the real

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