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**SHORT ANSWER TYPE QUESTIONS**

**Q1. Solve 2**x^{2} − 5x + 3 = 0.

**Sol**. We have:

2x^{2} − 5x + 3 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0,

∴ a = 2

b = − 5

c = 3

∴ b^{2} − 4ac =(− 5)^{2} − 4 (2) (3)

25 − 24 = 1

Since,

∴

Taking, + ve sign,

Taking, −ve sign,

Thus, the required roots are

x = 3/2 and x = 1**Q2. Solve the following quadratic equation: **

**x ^{2} + 4x − 8 = 0**

2x^{2 }+ 4x − 8= 0

Dividing by 2, we get

x^{2} + 2x − 4 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0,

a = 1

b = 2

c = − 4

∴ b^{2 }− 4ac = (2)^{2} − 4 (1) (− 4)

= 4 + 16 = 20

Since,

∴

⇒

Taking +ve sign, we get

Taking −ve sign we get,

Thus, the required roots are x = **Sol.** We have,

⇒ x^{2} + 3x + 2 + x^{2} − 3x + 2 =3 (x^{2} + x − 2)

⇒ 2x^{2} + 4 = 3x^{2} + 3x − 6

⇒ 3x^{2} + 3x − 6 − 2x^{2} − 4=0

⇒ x^{2} + 3x − 10 = 0

⇒ x^{2} + 5x − 2x − 10 = 0

⇒ x (x + 5) − 2 (x + 5) = 0

⇒ (x + 5) (x − 2) = 0

Either x + 5 = 0 ⇒ x = − 5

or x − 2 = 0 ⇒ x = 2

Thus, the required roots are

x = − 5 and x = 2**Q4. Solve (using quadratic formula): **

**x ^{2} + 5x + 5 = 0**

**Sol. **We have: x^{2} + 5x + 5 = 0

Comparing (1) with ax^{2} + bx + c = 0, we have:

a = 1

b = 5

c = 5

∴ b^{2} − 4ac = (5)^{2} − 4 (1) (5)

= 25 − 20 = 5

Since,

∴

⇒

Taking +ve sign, we have:

Taking −ve sign, we have:

Thus, the required roots are:

**Q5. Solve for x: 36x ^{2} − 12ax + (a^{2} − b^{2}) = 0.**

**Sol.** We have:

36x^{2} − 12ax + (a^{2} − b^{2}) = 0 ...(1)

Comparing (1) with Ax^{2} + Bx + C = 0, we have:

A = 36

B = − 12a

C = (a^{2} − b^{2})

∴ B^{2} − 4AC =[− 12a]^{2} − 4 (36) [a^{2} − b^{2}]

= 144 a^{2} − 144 (a^{2} − b^{2})

= 144 a^{2} − 144 a^{2} + 144 b^{2}

= 144 b^{2}

Taking +ve sign, we have:

⇒

Taking −ve sign, we get

⇒

Thus, the required roots are:

**Q6. Find the roots of the quadratic equation using the quadratic formula.**

**Sol.** Comparing the given equation with the general equation ax^{2} + bx + c = 0, we have

= 5 + 16

= 21

Now, using the quadratic formula, we have:

⇒

Taking the, positive sign, we get

Taking the negative sign, we get

**Q7. Solve: 16x ^{2} − 8a^{2} x + (a^{4} − b^{4}) = 0 for x.**

**Sol. **We have:

16x^{2} − 8a^{2} x + a^{4} − b^{4} = 0 ...(1)

Comparing (1) with ax^{2 }+ bx + c = 0, we get

a = 16

b = − 8a^{2}

c =(a^{4} − b^{4})

∴ b^{2} − 4ac =[− 8a^{2}]^{2} − 4 (16) (a^{4} − b^{4})

= 64 a^{4} − 64 (a^{4} − b^{4})

= 64 a^{4} − 64 a^{4} + 64 b^{4}

= 64 b^{4}

Since,

∴

⇒

Now, taking +ve sign, we get

Taking −ve sign, we get

Thus, the required roots are:

**Q8. Solve for x: 9x ^{2} − 6ax + a^{2} − b^{2} = 0.**

**Sol.** We have:

9x^{2} − 6ax + (a^{2 }− b^{2}) = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0, we get

a = 9, b = − 6a and c = (a^{2} − b^{2})

∴ b^{2} − 4ac = (− 6a)^{2 }− 4 (9) (a^{2} − b^{2})

= 36a^{2} − 36 (a^{2} − b^{2})

= 36a^{2} − 36a^{2} + 36b^{2}

= 36b^{2} = (6b)^{2}

Since,

∴

⇒

⇒

Taking the +ve sign, we get

Taking the −ve sign, we get

∴ The required roots are:

**Q9. Evaluate ****Sol. **

The given expression can be written as

or

squaring both side, we have

⇒ x^{2} = 20 + x

⇒ x^{2} – x – 20 = 0, where

Here : a =1, b = –1 and c = –20

∴

Since the given expression is positive,

∴ Rejecting the negative sign, we have:**Q10. Using quadratic formula, solve the following quadratic equation for x:**

x^{2} − 4ax + 4a^{2} − b^{2} = 0

**Sol**. Comparing the given equation with ax^{2} + bx + c = 0, we have:

a = 1

b = − 4a

c = 4a^{2} − b^{2}

∴ b^{2} − 4ac =[− (4a)]^{2} − 4 (1) [4a^{2} − b^{2}]

= 16a^{2} − 4 (4a^{2} − b^{2})

= 16a^{2} − 16a^{2} + 4b^{2}

= 4b2

= (2b)^{2}

Since,

∴

⇒

⇒

Taking the +ve sign, x = 2a + b

Taking the −ve sign, x = 2a − b

Thus, the required roots are:

x = 2a + b and x = 2a − b**Q11. Using quadratic formula, solve the following quadratic equation for x: **

**x ^{2} − 2ax + (a^{2} − b^{2}) = 0**

**Sol. **Comparing the given equation with ax^{2} + bx + c = 0, we have:

a = 1, b = − 2a and c = (a^{2} − b^{2})

∴ b^{2} − 4ac =(− 2a)^{2} − 4 (1) (a^{2} − b^{2})

= 4a^{2} − 4 (a^{2} − b^{2})

= 4a^{2} − 4a^{2} + 4b^{2} = 4b^{2}

Taking the +ve sign, we get

x = a + b

Taking the −ve sign, we get

x = a − b

Thus, the required roots are:

x = a + b and x = a − b**Q12. Solve for x : **

Sol. We have:

⇒

⇒

⇒

⇒

⇒ 2 × 3 = 2 (x − 1) (x − 3)

⇒ 3 = (x − 1) (x − 3)

⇒ x^{2} − 4x + 3 − 3 = 0

⇒ x^{2} − 4x = 0

⇒ x(x − 4) = 0

Either x = 0 or x = 4

Thus, x = 0 ; 4**Q13. Find the roots of the equation:**

**Sol.** We have:

⇒ (3x + 2) (7x + 9) = 11 (2x^{2} + 5x − 3)

⇒ 21x^{2} + 27x + 14x + 18 = 22x^{2} + 55x − 33

⇒ 21x^{2} + 41x + 18 = 22x^{2} + 55x − 33

⇒ (21 − 22) x^{2} + (41 − 55)x + (18 + 33) = 0

⇒ − x^{2} + (− 14x) + (51) = 0

⇒ x^{2} + 14x − 51 = 0

⇒ x^{2} + 17x − 3x − 51 = 0

⇒ x (x + 17) − 3 (x + 17) = 0

⇒ (x + 17) (x − 3) = 0

Either x + 17 =0 ⇒ x = − 17

or x − 3 = 0 ⇒ x = 3

Thus, the required roots of the given equation are:

3 and −17**Q14. Find the roots of the equation:**

**Sol.** We have:

⇒

⇒ − 3 × 3 = 4 × (x^{2} − 3x)

⇒ − 9 = 4x^{2} − 12x

⇒ 4x^{2} − 12x + 9 = 0 ...(1)

Comparing (1), with ax^{2} + bx + c = 0, we get

a = 4

b = − 12

c = 9

∴ b^{2} − 4ac =(− 12)^{2} − 4 (4) (9)

= 144 − 144 = 0

Now, the roots are:

⇒

⇒ x = 12/8 = 3/2

Thus, the roots are: 3/2 and 3/2**Q15. Solve for x :**

**Sol.**

⇒ 3 (x + 2) =(x − 2) (x + 1) or 3x + 6

= x^{2} − x − 2

⇒ x^{2} − 4x − 8 = 0

**Q16. Find the roots of the equation:**

**Sol.** We have:

⇒

⇒

⇒ (2x − 2) (2x + 5) = 8 (x^{2} − 2x)

⇒ 4x^{2} + 10x − 4x − 10 = 8x^{2} − 16x

⇒− 4x^{2} + 22x − 10 = 0

⇒ 2x^{2} − 11x + 5 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0, we have:

a = 2

b = − 11

c = 5

∴ b^{2} − 4ac = (− 11)^{2} − 4 (2) (5)

= 121 − 40 = 81

Now, the roots are given by

⇒

Taking the +ve sign,

Taking the −ve sign,

Thus, the required roots are: 5 and 1/2.**Q17. Solve : **

**Sol. **We have:

⇒ x (a + b + x) = −ab

⇒ x^{2 }+ (a + b)x + ab = 0

⇒ x^{2 } + ax + bx + ab = 0

⇒ x (x + a) + b (x + a) = 0

⇒ (x + a) (x + b) = 0

⇒ x + a = 0 or x + b = 0

∴ x = −a or x = −b**Q18. Find the roots of the following equation:**

**Sol.** We have:

⇒ − 11 × 30 = 11 × (x^{2} − 3x − 28)

⇒ − 30 = x^{2} − 3x − 28

⇒ x^{2} − 3x − 28 + 30 = 0

⇒ x^{2} − 3x + 2 = 0

⇒ x^{2} − 2x − x + 2 = 0

⇒ x (x − 2) − 1 (x − 2)= 0

⇒ (x − 1) (x − 2) = 0

Either x −1= 0 ⇒ x = 1

or x −2 = 0 ⇒ x = 2

Thus, the required roots are: 1 and 2.**Q19. If α and β are roots of the equation x ^{2} – 1 = 0, form an equation whose roots are **

**Sol.** ∵ α and β are roots of x^{2 }– 1 = 0 and x^{2} – 1 = 0 can be written as x^{2} + 0x – 1= 0 where a = 1, b = 0 and c = –1.

Now, the roots of the new equation are

∴ Sum of the roots of the new equation

Product of the roots of the new equation

Since, a quadratic equation is given by

∴ The required quadratic equation is

x^{2} – (–4) x + 4 = 0

or x^{2} + 4x + 4 = 0

**Remember** α^{2} + β^{2} + 2αβ =(α + β)^{2}

⇒ α^{2} + β^{2} =(α + β)^{2} – 2αβ

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