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3 years ago

Find the condition that one root of the quadratic equation may be 1 more than the other.

1 answer:

4 0

<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>

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What is the solution of 2|2x-1|-8=18?

Maurinko [17]

Solve for $|2x-1|$ :

$2|2x-1|-8=18\implies2|2x-1|=26\implies|2x-1|=13$

Now if $2x-1\ge0$ , we have $|2x-1|=2x-1$ and

$2x-1=13\implies2x=14\implies x=7$

On the other hand, if $2x-1$ , then $|2x-1|=-(2x-1)=1-2x$ and

$1-2x=13\implies-2x=12\implies x=-6$

4 0

3 years ago

Read 2 more answers

Solve for m 4=m+4 :) what does m = ?

Maslowich

Answer:

Step-by-step explanation:

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3 years ago

A chef cooks 9 lbs of chicken for 36 people and 17 lbs of chicken for 68 people

Anastaziya [24]

Answer:

$\frac{1}{4} lbs/person$

Step-by-step explanation:

Here is the complete question: Find the rate of change for the situation:

A chef cooks 9 lbs of chicken for 36 people and 17 lbs of chicken for 68 people.

Given: 1st situation, A chef cook 9 lbs chicken for 36 people.

2nd situation, chef cook 17 lbs chicken for 68 people.

∴ 1st situation, weight of chicken per person= $\frac{9\ lbs}{36\ person} = \frac{9}{36}$

Weight of chicken per person= $\frac{1}{4} lbs/ person$

2nd situation, weight of chicken per person= $\frac{17\ lbs}{68\ person} = \frac{17}{68}$

Weight of chicken per person= $\frac{1}{4} lbs/ person$

In both the situation chef cook same amount of chicken per person, which is $\frac{1}{4} lbs/ person$

∴ Rate of change is $\frac{1}{4} lbs/ person$

5 0

3 years ago

3,1,2,2,3,2,6,4 ayuda con esto

zalisa [80]

Answer:

Step-by-step explanation:

¿Qué necesitas para encontrar la mediana o el modo medio?

6 0

3 years ago

The population of a local species of dragonfly can be found using an infinite geometric series where a1=42 and the common ratio

san4es73 [151]

$\bf \textit{sum of an infinite geometric serie}\\\\
\stackrel{for~~|r|\ \textless \ 1}{S=\sum\limits_{i=0}^{\infty}~a_1r^i\implies \cfrac{a_1}{1-r}}\qquad 
\begin{cases}
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=42\\
r=\frac{3}{4}
\end{cases}
\\\\\\
S=\cfrac{42}{1-\frac{3}{4}}\implies S=\cfrac{42}{\frac{1}{4}}\implies S=164$

bearing in mind that, the geometric sequence is "convergent" only when |r|<1, or namely "r" is a fraction between 0 and 1.

3 0

3 years ago