BANK & INSURANCE (NEW PATTERN QUADRATIC EQUATION)

Since, one of the roots is -2, on putting a = -2 in the given equation the value obtained should be equal to '0'.

So, (-2)⁴ - b(-2)³ + 8 × (-2)² + 28 × (-2) - c = 0
Or, 16 + 8b + 32 - 56 - c = 0
Or, 8b - c = 8 .......... (I)

Similarly, on putting a = 3 in the given equation, we have;
(3)⁴ - b(3)³ + 8 × 3² + 28 × 3 - c = 0
Or, 81 - 27b + 72 + 84 - c = 0
Or, 27b + c = 237 .......... (II)

On adding equation (I) and (II), we have;
35b = 245
Or, b = 7
So, c = 7 × 8 - 8
So, c = 48

So, the given equation becomes a⁴ - 7a³ + 8a² + 28a - 48 = 0

For an equation of the form ax⁴ + bx³ + cx² + ex + d = 0
Sum of roots = (-b/a)
And, product of roots = (d/a)

Product of all four roots = -48

So, product of remaining two roots = (-48) ÷ (-2 × 3) = 8
And, sum of all four roots = 7
So, sum of other two roots = 7 + 2 - 3 = 6

1. : Let the other two roots be ‘x’ and ‘y’.
So, xy = 8 and x + y = 6
So, (x + y)² = x² + y² + 2xy
So, 6² = x² + y² + 2 × 8
Or, x² + y² = 20

Since, one of the roots is -2, on putting a = -2 in the given equation the value obtained should be equal to '0'.

So, (-2)⁴ - b(-2)³ + 8 × (-2)² + 28 × (-2) - c = 0
Or, 16 + 8b + 32 - 56 - c = 0
Or, 8b - c = 8 .......... (I)

Similarly, on putting a = 3 in the given equation, we have;
(3)⁴ - b(3)³ + 8 × 3² + 28 × 3 - c = 0
Or, 81 - 27b + 72 + 84 - c = 0
Or, 27b + c = 237 .......... (II)

On adding equation (I) and (II), we have;
35b = 245
Or, b = 7
So, c = 7 × 8 - 8
So, c = 48

So, the given equation becomes a⁴ - 7a³ + 8a² + 28a - 48 = 0

For an equation of the form ax⁴ + bx³ + cx² + ex + d = 0
Sum of roots = (-b/a)
And, product of roots = (d/a)

Product of all four roots = -48

So, product of remaining two roots = (-48) ÷ (-2 × 3) = 8
And, sum of all four roots = 7
So, sum of other two roots = 7 + 2 - 3 = 6

2. : Required value = 32 × 7 - 4 × 48 = 32

Since, one of the roots is -2, on putting a = -2 in the given equation the value obtained should be equal to '0'.

So, (-2)⁴ - b(-2)³ + 8 × (-2)² + 28 × (-2) - c = 0
Or, 16 + 8b + 32 - 56 - c = 0
Or, 8b - c = 8 .......... (I)

Similarly, on putting a = 3 in the given equation, we have;
(3)⁴ - b(3)³ + 8 × 3² + 28 × 3 - c = 0
Or, 81 - 27b + 72 + 84 - c = 0
Or, 27b + c = 237 .......... (II)

On adding equation (I) and (II), we have;
35b = 245
Or, b = 7
So, c = 7 × 8 - 8
So, c = 48

So, the given equation becomes a⁴ - 7a³ + 8a² + 28a - 48 = 0

For an equation of the form ax⁴ + bx³ + cx² + ex + d = 0
Sum of roots = (-b/a)
And, product of roots = (d/a)

Product of all four roots = -48

So, product of remaining two roots = (-48) ÷ (-2 × 3) = 8
And, sum of all four roots = 7
So, sum of other two roots = 7 + 2 - 3 = 6

3. : Given equation,
a² - (b + 1)a - c = 0
Or, a² - 8a - 48 = 0
Or, a² - 12a + 4a - 48 = 0
Or, a(a - 12) + 4(a - 12) = 0
Or, (a - 12)(a + 4) = 0
So, a = 12 or a = -4
Since, there’s only one positive root, required value = 12

p² - Pm + n = 0, have roots X and 5
Product of roots (5 × X) = n --------> (1)

p² - (m + 1)P + (n + 4) = 0, have roots (X + 2) and 4
Product of roots [(X + 2) × 4] = (n + 4) --------> (2)

From (1) and (2), we get
(X + 2) × 4 = 5X + 4
4X + 8 = 5X + 4
X = 4

So, value of n = 5 × 4 = 20
Sum of roots of equation I = (X + 5) = -(m)
Value of m = 5 + 4 = 9

4: 4% of 250 = 10

p² - Pm + n = 0, have roots X and 5
Product of roots (5 × X) = n --------> (1)

p² - (m + 1)P + (n + 4) = 0, have roots (X + 2) and 4
Product of roots [(X + 2) × 4] = (n + 4) --------> (2)

From (1) and (2), we get
(X + 2) × 4 = 5X + 4
4X + 8 = 5X + 4
X = 4

So, value of n = 5 × 4 = 20
Sum of roots of equation I = (X + 5) = -(m)
Value of m = 5 + 4 = 9

5: Required value = 4 + 9 + 20 = 33

For 1:
x² + 4x - 77 = 0
⇒ x = -11, 7

For 2:
y² - 22y + 105 = 0
⇒ y = 15, 7

For 3:
z² - 34z - 111 = 0
⇒ z² - 37z + 3z - 111 = 0
⇒ z(z - 37) + 3(z - 37) = 0
⇒ (z - 37)(z + 3) = 0
⇒ z = 37, -3

∴ The correct relation is 1- b, 2- c, 3- a

For eq. II ⇒ N(-13/2)² + 20(-13/2) - 52 × 4 = 0
N = 8

Now, 8r² + 20r - 52 × 4 = 0
4r² + 26r - 16r - 104 = 0
r = 4, -6.5

Given, the ratio of highest root of equation I to the highest root of equation II is 7:8

So, let the highest root of equation I = a
So, a/4 = 7/8
a = 7/2 = 3.5

For eq. I ⇒ M(7/2)² + 6(7/2) - 14 × 5 = 0
M = 4

Now, 4p² + 6p - 14 × 5 = 0
p = 3.5, -5

7: According to question, Required Ratio = (-5) × (-6.5) = 32.5

For eq. II ⇒ N(-13/2)² + 20(-13/2) - 52 × 4 = 0
N = 8

Now, 8r² + 20r - 52 × 4 = 0
4r² + 26r - 16r - 104 = 0
r = 4, -6.5

Given, the ratio of highest root of equation I to the highest root of equation II is 7:8

So, let the highest root of equation I = a
So, a/4 = 7/8
a = 7/2 = 3.5

For eq. I ⇒ M(7/2)² + 6(7/2) - 14 × 5 = 0
M = 4

Now, 4p² + 6p - 14 × 5 = 0
p = 3.5, -5

8: 4 + 3.5 = 7.5

From I: m² + 20m - 384 = 0

m² + 32m - 12m - 384 = 0
m(m + 32) - 12(m + 32) = 0
(m + 32)(m - 12) = 0
m = 12, -32

So, A = 12 and B = -32

From II:
n² + 9n - 532 = 0
n² + 28n - 19n - 532 = 0
n(n + 28) - 19(n + 28) = 0
(n + 28)(n - 19) = 0
n = 19, -28

So, C = 19 and D = -28

9: After putting values of ‘A’, ‘B’, ‘C’ and ‘D’ in the equation ‘III’, we get:

√{(D - B)² + (228A/C)} = ?

√{(-28 + 32)² + 228 × (12/19)} = ?
? = √{256 + (12 × 12)}
? = √(256 + 144)
? = √400
? = 20

From I: m² + 20m - 384 = 0

m² + 32m - 12m - 384 = 0
m(m + 32) - 12(m + 32) = 0
(m + 32)(m - 12) = 0
m = 12, -32

So, A = 12 and B = -32

From II:
n² + 9n - 532 = 0
n² + 28n - 19n - 532 = 0
n(n + 28) - 19(n + 28) = 0
(n + 28)(n - 19) = 0
n = 19, -28

So, C = 19 and D = -28

10: After putting values of ‘A’, ‘B’, ‘C’ and ‘D’ in the equation ‘III’, we get;

(C - A)⁵ - 37.5% of 4BD = ?

(13.5 - 10.5)⁵ - {37.5% × (4 × 9 × 12)} = ?

? = 3⁵ - (0.375 × 432)
? = 243 - 162
? = 81