Algebra

The quadratic formula x = (-b ± √(b²-4ac))/2a is correctly known as Shreedharacharya's formula, named after the ancient Indian mathematician who discovered it. Pythagoras is associated with the Pythagorean theorem, not quadratic equations. The other options are fictional names.

For ax²+bx+c: sum of roots = -b/a, product = c/a. Sum of reciprocals = (sum)/(product) = (-b/a)/(c/a) = -b/c = 22. For cx²+bx+a: product = a/c = 11, so a = 11c. From -b/c = 22, b = -22c. In ax²+bx+c: difference of roots = √[(sum)² - 4(product)] = √[(-b/a)² - 4(c/a)] = √[(22c/11c)² - 4(c/11c)] = √[(2)² - 4/11] = √[4 - 4/11] = √(40/11).

If 1.5 is a root, substitute x = 1.5 into x² + mx + 24 = 0: (1.5)² + m(1.5) + 24 = 0, giving 2.25 + 1.5m + 24 = 0, so 1.5m = -26.25, therefore m = -26.25/1.5 = -17.5. Verification: x² - 17.5x + 24 = 0 has roots 1.5 and 16 (product = 24, sum = 17.5).

For the quadratic equation 2x^2 + 2(p+1)x + p = 0, the discriminant is D = 4(p+1)^2 - 8p = 4(p^2+2p+1-2p) = 4(p^2+1). Since p^2+1 is always positive for all real p, D > 0 always. Therefore, the roots are always real and distinct.

First equation, x² - 4x + A = 0 Discriminant D = 16 - 4A Roots are (4 - (16 - 4A)^1/2)/2 and (4 + (16 - 4A)^1/2)/2, i.e. the roots are (2-(4-A)^1/2)and (2+(4-A)^1/2). Obviously, q = (2+(4-A)^1/2) and p = (2-(4-A)^1/2) because it is written that q>p. So, q-p = 2(4-A)^1/2 Second equation, x² - 12x + B =0 Discriminant D = 144 - 4B Roots are (12 - (144 - 4B)^1/2)/2 and (12 + (144 - 4B)^1/2)/2, i.e. the roots are (6-(36-B)^1/2)and (6+(36-B)^1/2). Obviously s = (6+(36-B)^1/2) and r = (6-(36-B)^1/2) because it is written that s>r. So, s-r = 2(36-B)^1/2 Since p, q, r and s are in AP, therefore q - p = s - r (In AP, common difference is the same) 2(4-A)^1/2 = 2(36-B)^1/2 Solving, we get 4-A = 36-B, i.e. B-A = 32 Now, possible combinations of (B,A) are (36,4), (35,3), (34,2), (33,1) because A<=4 as per the roots of first equation {(4-A)^1/2}, otherwise roots will be complex. Also, B<=36 as per the roots of the second equation. Hence, this option is correct.  

x³-3x²+3x+7=0 (x-1)³+8=0 (x-1)³=-8 (x-1)³=(-2)³ {(x-1)/(-2)}³=(1) Taking the cube root, we get (x-1)/(-2)=1,ω,ω² Solving, we get three different values of x or α, β, γ = -1, 1-2ω, 1-2ω² Putting the values of α, β, γ in the asked equation, (1/ω)+(1/ω)+ω²=3ω² (As 1 can be written as ω³)(Correct Answer)

Discriminant = b² - 4ac = 0 

It is the correct answer.

.b² - 4ac < 0 81 - 4m < 0 m > 20.25

Sum of roots = (a + b)² + (a - b)² = 2(a² + b²) = 2[(a + b)² - 2ab] = 2(49 - 24) = 50. Product = (a + b)²(a - b)² = 49(49 - 48) = 49

Given root 5, divide (x^3 - 10x^2 + 31x - 30) by (x-5) to get (x^2 - 5x + 6). Factoring (x^2 - 5x + 6) gives (x-2)(x-3). Thus, the roots are 2 and 3.

The polynomial will be of the form a((x - (1/4))(x +1)) On simplifying, and putting a = 4, we get the required polynomial as  4x² + 3x - 1 Hence, option (1 )is correct.

$\begin{array}{l} \left( { 5+\sqrt { 2 }  } \right) { x^{ 2 } }-\left( { 4+\sqrt { 5 }  } \right) x+8+2\sqrt { 5 } =0 \ a=5+\sqrt { 2 }  \ b=-\left( { 4+\sqrt { 5 }  } \right)  \ c=8+2\sqrt { 5 }  \ Harmonic\, \, mean\, \, of\, \, \lambda ,\beta  \ =\frac { { 2\lambda \beta  } }{ { \lambda +\beta  } }  \ \lambda \beta =\frac { c }{ a } =\frac { { 8+2\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ \lambda +\beta =\frac { { -b } }{ a } =\frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ Harmonic\, \, mean=\,  \ \frac { { 2\frac { { \left( { 8+2\sqrt { 5 }  } \right)  } }{ { 5\sqrt { 2 }  } }  } }{ { \frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  } }  \ =\frac { { 2\left( { 8+2\sqrt { 5 }  } \right)  } }{ { 4+\sqrt { 5 }  } }  \ =4\, \, \,  \end{array}$