1)The equation 2x²+ kx+3=0 has two equal roots , then the value of k is ??
2)If the roots of px²+qx+2=0 are reciprocal of each other ??
3)The roots of the quadratic equation 1/(a+b+x) = 1/a+1/b+1/x is??
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Answers
Question 1:-
➡ The equation 2x² + kx + 3 = 0 has two equal roots. What is the value of k?
Answer:-
➡ The value of k is 2√6
Solution:-
Given equation,
➡ 2x² + kx + 3 = 0
On comparing with ax² + bx + c = 0, we get,
➡ a = 2
➡ b = k
➡ c = 3
Now, it's given that the roots of the equation are equal.
So,
Discriminant = 0
➡ b² - 4ac = 0
➡ k²- 4×2×3 = 0
➡ k² - 24 = 0
➡ k² = 24
➡ k = √24
➡ k = 2√6
Hence, the value of k is 2√6
Question 2:-
If the roots of px² + qx + 2 = 0, are reciprocal of each other then?
1. p = 0
2. p = -2
3. p = ±2
4. p = 2
Answer:-
➡ Correct Option is 4.(i.e., p = 2)
Solution:-
Given equation,
px² + qx + 2 = 0
We know that,
➡ Product of roots = c/a
Since, the roots are reciprocal of each other,
So,
c/a = 1
➡ 2/p = 1
➡ p = 2
Hence, option 4 is the correct answer.
Question 3:-
➡ Find the roots of the education 1/(a+b+x) = 1/a + 1/b + 1/x (x ≠ 0, a ≠ 0, b ≠ 0)
Answer:-
➡ The roots of the equation are -a and -b
Solution:-
Given,
1/(a + b + x) = 1/a + 1/b + 1/x
➡ 1/(a + b + x) - 1/x = 1/a + 1/b
➡ (x - (a + b + x))/(x(a + b + x)) = (a + b)/ab
➡ -(a + b)/(x(a + b + x)) = (a + b)/ab
➡ -1/(x(a + b + x)) = 1/ab
➡ -ab = x(a + b + x)
➡ -ab = x(a + b) + x²
➡ x² + x(a + b) + ab = 0
On comparing with Ax² + Bx + C = 0, we get,
➡ A = 1
➡ B = (a + b)
➡ C = ab
Now,
x = (-b ± √(b² - 4ac))/2a
➡ x = (-a-b ± √((a+b)² - 4ab)/2
➡ x = (-a - b ± √(a² + b² + 2ab - 4ab))/2
➡ x = (-a - b ±√(a² + b² - 2ab))/2
➡ x = (-a - b ±(a - b))/2
So,
➡ x = (-a - b + a - b)/2 and x = (-a - b - a + b)/2
➡ x = -2b/2
➡ x = -b ......(i)
Also,
➡ x = (-a - b - a + ;)/2
➡ x = -2a/2
➡ x = -a ......(ii)
Hence, the roots of the equation are -a and -b.