Let base of the right triangle be x cm.
Given:

Perpendicular x+29 = 70

Perpendicular = (41−x) cm
Now, using Pythagoras theorem,
(29)²= x²+(41−x)2
841 = 1681+ x² −82x+ x²
2 x²−82x+840 = 0
x²− 41x+ 420 = 0
x²− 20x−21x+420 = 0
x(x−20)−21(x−20) = 0
(x−20)(x−21) = 0
x – 20 = 0 and x−21=0
x = 20 and x = 21
Therefore, the two sides other than hypotenuse are of 20 cm and 21 cm.
∴ Area of right triangle = 1/2 x (Base x Perpendicular)

= 1/2 x (20 x 21)

= 210 sq. cm

Using factorisation method
9a² b² x² −16abcdx−25c²d² = 0
9a²b²x² − 25abcdx + 9abcdx − 25c²d² = 0
abx(9abx−25cd) + cd(9abx−25cd) = 0
(abx+cd)(9abx−25cd) = 0

abx+cd = 0 and 9abx−25cd = 0

Let the number of wickets taken by Srinath be x
then the number of wickets taken by Kumble will be 2x−3
According to question, x(2x−3) = 20
2 x²−3 x−20 = 0
2 x²−8 x+5 x−20 = 0
2²(x−4) + 5(x−4) = 0
(x−4)(2x+5) = 0
x−4 = 0 and 2x+5 = 0
x = 4 and x = −52 [x = −52 is not possible]
Therefore, number of wickets taken by Srinath is 4.
Then number of wickets taken by Kumble = 2×4−3 = 5

If ax² + bx + c = 0 has equal roots, then
b²− 4ac = 0
4ac = b²

c = b²/4a

Since, If ax²+ bx + c = 0 has equal roots, then
b²− 4ac = 0
(a+c)² − 4ac = 0 [Given: b = a+c]
a² + c² + 2ac − 4ac = 0
a² + c² −2ac = 0
(a − c)² = 0
a – c = 0
a = c

If α and β are the roots of ax²+ bx + c = 0, then

A quadratic equation ax²+ bx + c = 0 has real and distinct roots, if b² − 4ac > 0

Discriminant of the quadratic equation ax²+ bx + c = 0 is D = b² − 4ac.

Given: (x ²+1) ² − x² = 0

(x²)²+2.x²+1-x²=0
(x² )²+2.x²-x²+1=0
(x² )²+x²+1=0
y=x² (Let)
y²+y+1=0

Determinant is given as

For the equation determinant is

Since, determinant is less than 0, y cannot have real value

Also, y=x² so, x also cannot have real values.

In equation x²+x−5 = 0
a =1, b =1, c = −5
∴ b² − 4acb² − 4ac = (1)²− 4×1×(−5)(1) = 1 + 20 = 21
Since b²−4ac >0

Therefore, x² + x – 5 = 0 has two distinct roots.