Question

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Answer 1

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Answer 2

$\displaystyle \mathtt{5.~a)~x^2-14x+15=0}\\ \\ \mathtt{a=1,~b=-14,~c=15}\\ \\ \mathtt{\mathbf{\Delta=b^2-4ac}=(-14)^2-4 \cdot 1 \cdot 15=196-60=136\ \textgreater \ 0}\\ \\ \mathtt{\mathsf{\Delta \ \textgreater \ 0~ecua\c{t}ia~are~dou\u{a}~solu\c{t}ii~reale~distincte: }}\\ \\ \mathtt{\mathbf{x_1= \frac{-b- \sqrt{\Delta} }{2a};~x_2= \frac{-b+ \sqrt{\Delta} }{2a}}}$

$\displaystyle \mathtt{x_1= \frac{-(-14)- \sqrt{136} }{2 \cdot 1}= \frac{14-2 \sqrt{34} }{2}= \frac{\not2(7- \sqrt{34} )}{\not2}=7- \sqrt{34}}\\ \\ \mathtt{x_2= \frac{-(-14)+ \sqrt{136} }{2 \cdot 1}= \frac{14+2 \sqrt{34} }{2}= \frac{\not2(7+ \sqrt{34}) }{\not2} =7+ \sqrt{34} }$

$\displaystyle \mathtt{b)~3x^2-4x+5=0}\\ \\ \mathtt{a=3,~b=-4,~c=5}\\ \\ \mathtt{\mathbf{\Delta=b^2-4ac}=(-4)^ 2-4 \cdot 3 \cdot 5=16-60=-44\ \textless \ 0}\\ \\ \mathtt{\mathsf{\Delta\ \textless \ 0~ecua\c{t}ia~nu~are~solu\c{t}ii~reale~dar~are~dou\u{a}~solu\c{t}ii~complexe:}}\\ \\ \mathtt{\mathbf{x_1= \frac{-b-i \sqrt{-\Delta} }{2a};~x_2= \frac{-b+i \sqrt{-\Delta} }{2a}}}$

$\displaystyle \mathtt{x_1= \frac{-(-4)-i \sqrt{44}}{2 \cdot3}= \frac{4-2 \sqrt{11}i }{6}= \frac{\not2(2- \sqrt{11}i)}{\not6~~_3}= \frac{2- \sqrt{11}i }{3}=}\\ \\ \mathtt{= \frac{2}{3}- \frac{ \sqrt{11} }{3} i }\\ \\ \mathtt{x_2= \frac{-(-4)+i \sqrt{44}}{2 \cdot 3}= \frac{4+2 \sqrt{11}i}{6}= \frac{\not2(2+ \sqrt{11}i)}{\not6~~_3}= \frac{2+ \sqrt{11}i}{3}=}\\ \\ \mathtt{= \frac{2}{3}+ \frac{ \sqrt{11} }{3}i}$

$\displaystyle \mathtt{c)~5x^2-8x+3=0}\\ \\ \mathtt{a=5,~b=-8,~c=3}\\ \\ \mathtt{\mathbf{\Delta=b^2-4ac}=(-8)^2-4 \cdot 5 \cdot 3=64-60=4\ \textgreater \ 0}\\ \\ \mathtt{\mathsf{\Delta\ \textgreater \ 0\Rightarrow \mathbf{x_1= \frac{-b- \sqrt{\Delta} }{2a} ;~x_2= \frac{-b+ \sqrt{\Delta} }{2a} }}}$

$\displaystyle \mathtt{x_1= \frac{-(-8)- \sqrt{4}}{2 \cdot 5}= \frac{8-2}{10}= \frac{6}{10}= \frac{3}{5}}\\ \\ \mathtt{x_2= \frac{-(-8)+ \sqrt{4} }{2 \cdot 5}= \frac{8+2}{10}= \frac{10}{10}=1 }$

$\displaystyle \mathtt{d)~-3x^2+4x-1=0|\cdot (-1)}\\ \\ \mathtt{3x^2-4x+1=0}\\ \\ \mathtt{a=3,~b=-4,~c=1}\\ \\ \mathtt{\mathbf{\Delta=b^2-4ac}=(-4)^2-4 \cdot 3 \cdot 1=16-12=4\ \textgreater \ 0}\\ \\ \mathtt{\mathsf{\Delta \ \textgreater \ 0 \Rightarrow \mathbf{x_1= \frac{-b- \sqrt{\Delta} }{2a};~x_2= \frac{-b+ \sqrt{\Delta} }{2a}}}}$

$\displaystyle \mathtt{x_1= \frac{-(-4)- \sqrt{4} }{2 \cdot 3}= \frac{4-2}{6}= \frac{2}{6}= \frac{1}{3}}\\ \\ \mathtt{x_2= \frac{-(-4)+ \sqrt{4} }{2 \cdot 3}= \frac{4+2}{6}= \frac{6}{6}=1}$

$\displaystyle \mathtt{6.~a)~3x^2-5x+2=0}\\ \\ \mathtt{a=3,~b=-5,~c=2}\\ \\ \mathtt{x_1+x_2=- \frac{-5}{3}= \frac{5}{3} }\\ \\ \mathtt{x_1\cdot x_2= \frac{2}{3} }\\ \\ \mathtt{b)~-2x^2+6x-16=0}\\ \\ \mathtt{a=-2,~b=6,~c=-16}\\ \\ \mathtt{x_1+x_2=- \frac{6}{-2}=3}\\ \\ \mathtt{x_1 \cdot x_2= \frac{-16}{-2}=8}\\ \\ \mathtt{c)~3x^2-10x-2=0}\\ \\ \mathtt{a=3,~b=-10,~c=-2}\\ \\ \mathtt{x_1+x_2=- \frac{-10}{3} = \frac{10}{3} }\\ \\ \mathtt{x_1 \cdot x_2= -\frac{2}{3} }$