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Meghna Thapar 5 years, 6 months ago
A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable. The Babylonians came up with a technique called “completing the square” to solve common problems with areas by 400 BC. The first purely mathematical try to come up with a quadratic formula was done by Pythagoras in 500 BC. Euclid did the same thing in Alexandria, Egypt. Euclid used a purely geometric method.
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Manvi Jain 5 years, 6 months ago
Yogita Ingle 5 years, 6 months ago
We have two Equations:
2x +3y= 5
6x +ky= 15
For No Solution, We have:
a1/a2 = b1/b2 ≠ c1/c2
Here, a1 = 2 , a2 = 6 , c2 = 15 , c1 =5 & b1 = 3 , b2 = k
Substitute given values in Formula:
a1/a2 = b1/b2 ≠ c1/c2
→ 2/6 = 3/k ≠ 5/15
→ 2/6 = 3/k
→ 18 = 2k
→ k = 18/2
→ k = 9
Therefore, Value of k is 9.
Posted by Sakshi Khilari 5 years, 6 months ago
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Rajan Kumar Pasi 5 years, 6 months ago
Sol: Let the speed of the train = x kmph and that of the car = y kmph.
Total distance = 400 kms.
Time 1 = t1 = 4 hrs 30 mins = 4.5 hrs.
Time 2 = t2 = 4 hrs 30 mins + 15 mins = 4 hrs 45 mins = 4.75 hrs.
Physics formula used = Distance = speed x time.
Case 1:
If he covers half the distance by train and the rest of by the car then he travels 200 kms by each.
total time taken = distance/speed of train + distance/speed of car
{tex}\huge {=> \ 4.5= \frac {200}{x} + \frac {200}{y}}\\
\huge {=> \ \frac{4.5}{200}= \frac {1}{x} + \frac {1}{y}}\\
\text{Let u = }{\frac1x} \text{ and v = }{\frac1y}\\
\huge {=> \ \{\frac{4.5}{200}= u+v}\}.............eq.1\\{/tex}
Case 2:
if he travels 100km by train and rest by the car, then he travels 300 kms by car.
total time taken = distance/speed of train + distance/speed of car
{tex}\huge {=> \ 4.75= \frac {100}{x} + \frac {300}{y}}\\
\huge {=> \ \{\frac{4.75}{100}= u + 3v}\}.............eq.2\\{/tex}
Subtracting eq. 2 from eq. 1, we will get:
{tex}\huge {=> \ \frac{4.75}{100}-\frac{4.5}{200}= 2v}\\ \huge {=> \ \frac{1}{40}= 2v}\\ \huge {=> \ v=\frac{1}{80}}\\ \text{Therefore, putting value of v in eq. 1 we will get:}\\ \huge {=> \ u=\frac{1}{100}}\\{/tex}
{tex}\huge {=> \ u=\frac1x=\frac{1}{100}}\\ .\\ \text{Therefore, } \large \boxed{x=100} \ \ and,\\ \huge {=> \ v=\frac1x=\frac{1}{80}}\\ .\\ \text{Therefore, } \large \boxed{y=80}\\ \text{Therefore, } \large \text{ speed of the train = 100 kmph and speed of the car = 80 kmph.}\\ {/tex}
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Rajan Kumar Pasi 5 years, 6 months ago
if you have four zeroes, let say {tex}\huge \alpha, \beta,\gamma,\delta{/tex}
Then it belongs to a bi-quadratic or 4th power equation.
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Rajan Kumar Pasi 5 years, 6 months ago
{tex}\huge {=>\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}}\\ \huge {=>\ x = {-(-2) \pm \sqrt{(-2)^2-4(1)(8)} \over 2\times 1}}\\ \huge {=>\ x = {2 \pm \sqrt{4-32} \over 2}}\\ \huge {=>\ x = {2 \pm \sqrt{-28} \over 2}}\\ \huge {=>\ x = {1 \pm \sqrt{-7} }}\\ \text{Since,} \huge{ {b^2-4ac}}\ is\ smaller\ than\ 0.\ Thus,\ no\ real\ roots\ or\ zeros\ exist.{/tex}
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Anushka ???? 5 years, 6 months ago
5Thank You