Question 1209247
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Find the constant p such that x^2 - 5x - 14x + x^2 + p is the square of a binomial.
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<pre>
Reduce to the standard form quadratic polynomial by combining like terms

    2x^2 - 19x + p.


Present it as a square of a binomial

    2x^2 - 19x + p = (ax+b)^2.


It is the same as

    2x^2 - 19x + p = a^2*x^2 + 2abx + b^2.


Hence,

    2   = a^2,  or  a = {{{sqrt(2)}}};

    -19 = 2ab,  or  -19 = {{{2*sqrt(2)*b}}},  which gives  b = {{{-19/(2*sqrt(2))}}} = {{{-(19*sqrt(2))/4}}};

    p = b^2 = {{{361*2/16}}} = {{{361/8}}}.
</pre>

Solved.