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nignag [31]
3 years ago
7

Help please!! How do I solve this

Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
Check 2:   2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 => 
1-5+4=0 => 0=0  Identity provin solution.
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statuscvo [17]

Answer: The answer is \textup{The other root is }\dfrac{8}{3}~\textup{and}q=40.Step-by-step explanation:  The given quadratic equation is[tex]3x^2+7x-q=0\\\\\Rightarrow x^2-\dfrac{7}{3}x-\dfrac{q}{3}=0.

Also given that -5 is one of the roots, we are to find the other root and the value of 'q'.

Let the other root of the equation be 'p'. So, we have

p-5=-\dfrac{7}{3}\\\\\\\Rightarrow p=5-\dfrac{7}{3}\\\\\\\Rightarrow p=\dfrac{8}{3},

and

p\times(-5)=-\dfrac{q}{3}\\\\\\\Rightarrow \dfrac{8}{3}\times 5=\dfrac{q}{3}\\\\\\\Rightarrow q=40.

Thus, the other root is \dfrac{8}{3} and the value of 'q' is 40.

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Step-by-step explanation:

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Decreasing it by 20% then

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