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**Q1.** If two of the roots of f(x) = x^{3} – 5x^{2} – 16x + 80 are equal in magnitude but opposite in sign, then find all of its zeroes.

**Hint:** Let α and β are the two zeroes which are equal in magnitude but opposite in sign.

∴ α + β = 0

(Let the third zero is γ)

Sum of zeroes of

f(x)= α + β + γ =

∴ γ = 5 [α + β = 0]

Product of zeroes

∴

⇒ – α^{2} = –16 [α + β = 0 ⇒ β = – α]

⇒ α^{2} = 16 ⇒ α = ± 4

α = ± 4 ⇒ β = ∓ 4

[∴ β = –α]

Thus, the zeroes are : [± 4, ∓ 4, and 5]**Q2.** **Hint:**

Adding (1) and (2), we get x = 1/3 and y = 1/2**Q3.** Solve : **Hint:** Put x + 2y = p and 2x – y = q

We have ...(1)

...(2)

Solving (1) and (2), we get p = 4 and q = 3

∴ x + 2y = 4 and 2x – y = 3

Solving these equations, we get x = 2 and y = 1**Q4.** Solve : x + y = 18 ; y + z = 12 ; z + x = 16.**Hint: **Adding the three equations, we get

⇒ x + y + z = 23

Now, (x + y + z = 23) – (x + y = 18)

⇒ z = 5 (x + y + z = 23) – (y + z = 12)

⇒ x = 11 (x + y + z = 23) – (z + x = 16)

⇒ y = 7

Thus, x = 11, y = 7 and z = 5**Q5. **Solve : **Hint: **Inverting the equations:

Adding (1), (2) and (3), we get

Now, subtracting (1), (2) and (3) turn by turn from (4), we get x = 2, y = 4 and z = 6**Q6. **Solve: **Hint:**

From

⇒ 11(x + y – 3) = 2(3x + y) ⇒ 5x + 9y = 33 ...(1)

From

⇒ 11(x + 2y – 4) = 3(3x + y)

⇒ 2x + 9y = 44 ...(2)

Solving (1) and (2), we get x = 3 and y = 2

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