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3 years ago

Use the quadratic formula to solve the equation 25x^2-30x+25=0

Mathematics

1 answer:

Reika [66]3 years ago

8 0

Answer: $x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}$

Step-by-step explanation:

$25x^2-30x+25=0$

$x_{1,\:2}=\frac{-\left(-30\right)\pm \sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}}{2\cdot \:25}$

$\sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}$

$=\sqrt{30^2-4\cdot \:25\cdot \:25}$

$=\sqrt{30^2-2500}$

$=i\sqrt{2500-30^2}$

$=40i$

$x_{1,\:2}=\frac{-\left(-30\right)\pm \:40i}{2\cdot \:25}$

$x_1=\frac{-\left(-30\right)+40i}{2\cdot \:25},\:x_2=\frac{-\left(-30\right)-40i}{2\cdot \:25}$

$x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}$

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Step-by-step explanation:

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3 years ago

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Morgarella [4.7K]

Answer:

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Step-by-step explanation:

We can find the slope

m = (y2-y1)/ (x2-x1)

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3 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

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3 years ago