Решить дискриминант -t^2+5t-4. Created by rasila19042002. algebra-ru

Решить дискриминант -t^2+5t-4...

rubilovalex

$-t^2+5t-4=0\\\\D=b^2-4ac=5^2-4\cdot (-1)\cdot (-4)=25-16=9\; ,\; \; \underline {D=9}\\\\x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}= \frac{-5\pm \sqrt9}{2\cdot (-1)}=\frac{-5\pm 3}{-2}\\\\x_1=\frac{-5-3}{-2}=4\; ,\; \; x_2=\frac{-5+3}{-2}=1$

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Verified answer

-t^2+5t-4=0; D=5^2-4*(-1)*(-4)=25-16=9; t1=(-5-3)/(-2), t2=(-5+3)/(-2). t1=4, t2=1.

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