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**SHORT ANSWER TYPE QUESTIONS**

**Q1. Solve 2**x^{2} âˆ’ 5x + 3 = 0.

**Sol**. We have:

2x^{2} âˆ’ 5x + 3 = 0 ...(1)

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

âˆ´ a = 2

b = âˆ’ 5

c = 3

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 5)^{2} âˆ’ 4 (2) (3)

25 âˆ’ 24 = 1

Since,

âˆ´

Taking, + ve sign,

Taking, âˆ’ve sign,

Thus, the required roots are

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

**x ^{2} + 4x âˆ’ 8 = 0**

2x^{2 }+ 4x âˆ’ 8= 0

Dividing by 2, we get

x^{2} + 2x âˆ’ 4 = 0 ...(1)

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

a = 1

b = 2

c = âˆ’ 4

âˆ´ b^{2 }âˆ’ 4ac = (2)^{2} âˆ’ 4 (1) (âˆ’ 4)

= 4 + 16 = 20

Since,

âˆ´

â‡’

Taking +ve sign, we get

Taking âˆ’ve sign we get,

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

â‡’ x^{2} + 3x + 2 + x^{2} âˆ’ 3x + 2 =3 (x^{2} + x âˆ’ 2)

â‡’ 2x^{2} + 4 = 3x^{2} + 3x âˆ’ 6

â‡’ 3x^{2} + 3x âˆ’ 6 âˆ’ 2x^{2} âˆ’ 4=0

â‡’ x^{2} + 3x âˆ’ 10 = 0

â‡’ x^{2} + 5x âˆ’ 2x âˆ’ 10 = 0

â‡’ x (x + 5) âˆ’ 2 (x + 5) = 0

â‡’ (x + 5) (x âˆ’ 2) = 0

Either x + 5 = 0 â‡’ x = âˆ’ 5

or x âˆ’ 2 = 0 â‡’ x = 2

Thus, the required roots are

x = âˆ’ 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

âˆ´ b^{2} âˆ’ 4ac = (5)^{2} âˆ’ 4 (1) (5)

= 25 âˆ’ 20 = 5

Since,

âˆ´

â‡’

Taking +ve sign, we have:

Taking âˆ’ve sign, we have:

Thus, the required roots are:

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

**Sol.** We have:

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

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

A = 36

B = âˆ’ 12a

C = (a^{2} âˆ’ b^{2})

âˆ´ B^{2} âˆ’ 4AC =[âˆ’ 12a]^{2} âˆ’ 4 (36) [a^{2} âˆ’ b^{2}]

= 144 a^{2} âˆ’ 144 (a^{2} âˆ’ b^{2})

= 144 a^{2} âˆ’ 144 a^{2} + 144 b^{2}

= 144 b^{2}

Taking +ve sign, we have:

â‡’

Taking âˆ’ve sign, we get

â‡’

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:

â‡’

Taking the, positive sign, we get

Taking the negative sign, we get

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

**Sol. **We have:

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

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

a = 16

b = âˆ’ 8a^{2}

c =(a^{4} âˆ’ b^{4})

âˆ´ b^{2} âˆ’ 4ac =[âˆ’ 8a^{2}]^{2} âˆ’ 4 (16) (a^{4} âˆ’ b^{4})

= 64 a^{4} âˆ’ 64 (a^{4} âˆ’ b^{4})

= 64 a^{4} âˆ’ 64 a^{4} + 64 b^{4}

= 64 b^{4}

Since,

âˆ´

â‡’

Now, taking +ve sign, we get

Taking âˆ’ve sign, we get

Thus, the required roots are:

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

**Sol.** We have:

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

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

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

âˆ´ b^{2} âˆ’ 4ac = (âˆ’ 6a)^{2 }âˆ’ 4 (9) (a^{2} âˆ’ b^{2})

= 36a^{2} âˆ’ 36 (a^{2} âˆ’ b^{2})

= 36a^{2} âˆ’ 36a^{2} + 36b^{2}

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

Since,

âˆ´

â‡’

â‡’

Taking the +ve sign, we get

Taking the âˆ’ve sign, we get

âˆ´ The required roots are:

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

The given expression can be written as

or

squaring both side, we have

â‡’ x^{2} = 20 + x

â‡’ x^{2} â€“ x â€“ 20 = 0, where

Here : a =1, b = â€“1 and c = â€“20

âˆ´

Since the given expression is positive,

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

x^{2} âˆ’ 4ax + 4a^{2} âˆ’ b^{2} = 0

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

a = 1

b = âˆ’ 4a

c = 4a^{2} âˆ’ b^{2}

âˆ´ b^{2} âˆ’ 4ac =[âˆ’ (4a)]^{2} âˆ’ 4 (1) [4a^{2} âˆ’ b^{2}]

= 16a^{2} âˆ’ 4 (4a^{2} âˆ’ b^{2})

= 16a^{2} âˆ’ 16a^{2} + 4b^{2}

= 4b2

= (2b)^{2}

Since,

âˆ´

â‡’

â‡’

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

Taking the âˆ’ve sign, x = 2a âˆ’ b

Thus, the required roots are:

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

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

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

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

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 2a)^{2} âˆ’ 4 (1) (a^{2} âˆ’ b^{2})

= 4a^{2} âˆ’ 4 (a^{2} âˆ’ b^{2})

= 4a^{2} âˆ’ 4a^{2} + 4b^{2} = 4b^{2}

Taking the +ve sign, we get

x = a + b

Taking the âˆ’ve sign, we get

x = a âˆ’ b

Thus, the required roots are:

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

Sol. We have:

â‡’

â‡’

â‡’

â‡’

â‡’ 2 Ã— 3 = 2 (x âˆ’ 1) (x âˆ’ 3)

â‡’ 3 = (x âˆ’ 1) (x âˆ’ 3)

â‡’ x^{2} âˆ’ 4x + 3 âˆ’ 3 = 0

â‡’ x^{2} âˆ’ 4x = 0

â‡’ x(x âˆ’ 4) = 0

Either x = 0 or x = 4

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

**Sol.** We have:

â‡’ (3x + 2) (7x + 9) = 11 (2x^{2} + 5x âˆ’ 3)

â‡’ 21x^{2} + 27x + 14x + 18 = 22x^{2} + 55x âˆ’ 33

â‡’ 21x^{2} + 41x + 18 = 22x^{2} + 55x âˆ’ 33

â‡’ (21 âˆ’ 22) x^{2} + (41 âˆ’ 55)x + (18 + 33) = 0

â‡’ âˆ’ x^{2} + (âˆ’ 14x) + (51) = 0

â‡’ x^{2} + 14x âˆ’ 51 = 0

â‡’ x^{2} + 17x âˆ’ 3x âˆ’ 51 = 0

â‡’ x (x + 17) âˆ’ 3 (x + 17) = 0

â‡’ (x + 17) (x âˆ’ 3) = 0

Either x + 17 =0 â‡’ x = âˆ’ 17

or x âˆ’ 3 = 0 â‡’ x = 3

Thus, the required roots of the given equation are:

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

**Sol.** We have:

â‡’

â‡’ âˆ’ 3 Ã— 3 = 4 Ã— (x^{2} âˆ’ 3x)

â‡’ âˆ’ 9 = 4x^{2} âˆ’ 12x

â‡’ 4x^{2} âˆ’ 12x + 9 = 0 ...(1)

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

a = 4

b = âˆ’ 12

c = 9

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 12)^{2} âˆ’ 4 (4) (9)

= 144 âˆ’ 144 = 0

Now, the roots are:

â‡’

â‡’ x = 12/8 = 3/2

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

**Sol.**

â‡’ 3 (x + 2) =(x âˆ’ 2) (x + 1) or 3x + 6

= x^{2} âˆ’ x âˆ’ 2

â‡’ x^{2} âˆ’ 4x âˆ’ 8 = 0

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

**Sol.** We have:

â‡’

â‡’

â‡’ (2x âˆ’ 2) (2x + 5) = 8 (x^{2} âˆ’ 2x)

â‡’ 4x^{2} + 10x âˆ’ 4x âˆ’ 10 = 8x^{2} âˆ’ 16x

â‡’âˆ’ 4x^{2} + 22x âˆ’ 10 = 0

â‡’ 2x^{2} âˆ’ 11x + 5 = 0 ...(1)

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

a = 2

b = âˆ’ 11

c = 5

âˆ´ b^{2} âˆ’ 4ac = (âˆ’ 11)^{2} âˆ’ 4 (2) (5)

= 121 âˆ’ 40 = 81

Now, the roots are given by

â‡’

Taking the +ve sign,

Taking the âˆ’ve sign,

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

**Sol. **We have:

â‡’ x (a + b + x) = âˆ’ab

â‡’ x^{2 }+ (a + b)x + ab = 0

â‡’ x^{2 } + ax + bx + ab = 0

â‡’ x (x + a) + b (x + a) = 0

â‡’ (x + a) (x + b) = 0

â‡’ x + a = 0 or x + b = 0

âˆ´ x = âˆ’a or x = âˆ’b**Q18. Find the roots of the following equation:**

**Sol.** We have:

â‡’ âˆ’ 11 Ã— 30 = 11 Ã— (x^{2} âˆ’ 3x âˆ’ 28)

â‡’ âˆ’ 30 = x^{2} âˆ’ 3x âˆ’ 28

â‡’ x^{2} âˆ’ 3x âˆ’ 28 + 30 = 0

â‡’ x^{2} âˆ’ 3x + 2 = 0

â‡’ x^{2} âˆ’ 2x âˆ’ x + 2 = 0

â‡’ x (x âˆ’ 2) âˆ’ 1 (x âˆ’ 2)= 0

â‡’ (x âˆ’ 1) (x âˆ’ 2) = 0

Either x âˆ’1= 0 â‡’ x = 1

or x âˆ’2 = 0 â‡’ x = 2

Thus, the required roots are: 1 and 2.**Q19. If Î± and Î² are roots of the equation x ^{2} â€“ 1 = 0, form an equation whose roots are **

**Sol.** âˆµ Î± and Î² are roots of x^{2 }â€“ 1 = 0 and x^{2} â€“ 1 = 0 can be written as x^{2} + 0x â€“ 1= 0 where a = 1, b = 0 and c = â€“1.

Now, the roots of the new equation are

âˆ´ Sum of the roots of the new equation

Product of the roots of the new equation

Since, a quadratic equation is given by

âˆ´ The required quadratic equation is

x^{2} â€“ (â€“4) x + 4 = 0

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

**Remember** Î±^{2} + Î²^{2} + 2Î±Î² =(Î± + Î²)^{2}

â‡’ Î±^{2} + Î²^{2} =(Î± + Î²)^{2} â€“ 2Î±Î²

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