Question

$2x {}^{2} + x sqrt{5} - 15 =0$
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Answer 1

Ответ:

Объяснение:

$[begin{gathered}2{x^2}+xsqrt5-15=0hfill\D={b^2}-4ac={left({sqrt5}right)^2}-4cdot2cdot(-15)=5+120=125hfill\{x_{1;2}}=frac{{-bpmsqrt D}}{{2a}}=frac{{-sqrt5pmsqrt{125}}}{{2cdot2}}=frac{{-sqrt5pm5sqrt5}}{4}hfill\{x_1}=frac{{-sqrt5+5sqrt5}}{4}=frac{{sqrt5(-1+5)}}{4}=frac{{sqrt5cdot4}}{4}=sqrt5hfill\{x_2}=frac{{-sqrt5-5sqrt5}}{4}=frac{{sqrt5(-1-5)}}{4}=frac{{sqrt5cdot(-6)}}{4}=-frac{{3sqrt5}}{2}hfill\end{gathered}]$

Ответ: $displaystyle [{x_1}=sqrt 5;;{x_2}=-frac{{3sqrt 5}}{2}]$