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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In the diagram below, the area of $\triangle BCD$ is 60. What is the length of side $\overline{BC}$?

Phantasy [73]

Area of triangle = 1/2 bh

60 = 1/2 (15) h

8 = h

Quadrilateral area = h x (b1 + b2)/2 = 8 x (15+9)/2 = 96 sq units

96 -60 = triangle BAD area = 36 sq units

8 0

3 years ago

Pls help me answer this question thx

nadya68 [22]

Answer:

$\large\boxed{\dfrac{1}{115}}$

Step-by-step explanation:

$\text{Use}\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^{-1}=\dfrac{1}{a}\\--------------------------\\\\\dfrac{5^{2n-1}-4(25^{n-1})}{(5^n\cdot2)^2+3(5^{2n-1})}=\dfrac{5^{2n}\cdot5^{-1}-4\left[(5^2)^n\cdot25^{-1}\right]}{(5^n)^2\cdot2^2+3\left[(5^2)^n\cdot5^{-1}\right]}$

$=\dfrac{5^{2n}\cdot\frac{1}{5}-4\left(5^{2n}\cdot\frac{1}{25}\right)}{5^{2n}\cdot4+3\left(5^{2n}\cdot\frac{1}{5}\right)}=\dfrac{5^{2n}\left(\frac{1}{5}-\frac{4}{25}\right)}{5^{2n}\left(4+\frac{3}{5}\right)}=\dfrac{\frac{5}{25}-\frac{4}{25}}{4\frac{3}{5}}=\dfrac{1}{25}:\left(\dfrac{4\cdot5+3}{5}\right)\\\\=\dfrac{1}{25}\cdot\dfrac{5}{23}=\dfrac{1}{5}\cdot\dfrac{1}{23}=\dfrac{1}{115}$