Q: √ 6 + √6 + √6 +........ is equal to

As per the given above question , we have Let, x = 6 + √6 + √6 + √6 +........ On squaring both sides, we have x2 = 6 + √6 + √6 + √6 +........ ⇒ x2 = 6 + x ⇒ x2 - x - 6 = 0 ⇒ x2 - 3x + 2x - 6 = 0 ⇒ x( x - 3 ) + 2( x - 3 ) = 0 ⇒ ( x - 3 ) ( x + 2 ) = 0 ⇒ x = 3 and x ≠ −2 because numbers are positive. Thus , correct answer will be 3 .

Q: Form a quadratic equation whose one root is 3 - √5 and the sum of roots is 6

Given : - Sum of the roots ( k1 + k2 ) = 6 One root k1 = 3 - √5 ∴ the other root k2 = 6 - ( 3 - √5) = 3 + √5 ∴ Product of roots k1k2 = ( 3 - √5 ) ( 3 + √5 ) product of roots = 9 - 5 = 4 Hence , the required equation is x2 - (sum of the roots)x + (product of roots) = 0 ⇒ x2 - 6x + 4 = 0.

Q: Construct a quadratic equation whose roots are √2 and 2 √2

Let k1 and k2 be the roots of quadratic equation . Given :- k1 = √2 and k2 = 2 √2 Sum of the roots = √2 + 2 √2 = 3 √2 Product of the roots = ( √2 ) ( 2√2 ) = 4 Hence the required quadratic equation is x 2 = (sum of the roots) x + (Product of two roots) = 0 ⇒ x 2 - 3 √2x + 4 = 0.

Question 1

Ⅰ. ( 17 ) 2 + 144 ÷ 18 = x Ⅱ. ( 26 ) 2 - 18 × 21 = y

Accepted Answer

Question 2

Ⅰ. x 3 × 13 = x 2 × 247 Ⅱ. y 1/3 × 14 = 294 ÷ y 2/3

Accepted Answer

Question 3

Ⅰ. √ 500x + √ 402 = 0 Ⅱ. √ 360y + ( 200 ) 1/2 = 0

Accepted Answer

As per the given above equations , we can say that Ⅰ. √500x + √402 = 0 ⇒ √500x = - √402 ∴ x = - √( 402/500 ) = - √( 400/500 ) = - 0.9 Ⅱ. √360y + ( 200 )1/2 = 0 ⇒ √360y = - √200 ∴ y = - √( 200 / 360 ) = - 0.74 Clearly, we can see that x < y is required answer .

Question 4

√ 6 + √6 + √6 +........ is equal to

Accepted Answer

As per the given above question , we have Let, x = 6 + √6 + √6 + √6 +........ On squaring both sides, we have x2 = 6 + √6 + √6 + √6 +........ ⇒ x2 = 6 + x ⇒ x2 - x - 6 = 0 ⇒ x2 - 3x + 2x - 6 = 0 ⇒ x( x - 3 ) + 2( x - 3 ) = 0 ⇒ ( x - 3 ) ( x + 2 ) = 0 ⇒ x = 3 and x ≠ −2 because numbers are positive. Thus , correct answer will be 3 .

Question 5

Form a quadratic equation whose one root is 3 - √5 and the sum of roots is 6

Accepted Answer

Given : - Sum of the roots ( k1 + k2 ) = 6 One root k1 = 3 - √5 ∴ the other root k2 = 6 - ( 3 - √5) = 3 + √5 ∴ Product of roots k1k2 = ( 3 - √5 ) ( 3 + √5 ) product of roots = 9 - 5 = 4 Hence , the required equation is x2 - (sum of the roots)x + (product of roots) = 0 ⇒ x2 - 6x + 4 = 0.

Question 6

Determine p so that the equation x 2 + 5px + 16 = 0 has on real root.

Accepted Answer

Question 7

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

Accepted Answer

Question 8

Construct a quadratic equation whose roots are √2 and 2 √2

Accepted Answer

Let k1 and k2 be the roots of quadratic equation . Given :- k1 = √2 and k2 = 2 √2 Sum of the roots = √2 + 2 √2 = 3 √2 Product of the roots = ( √2 ) ( 2√2 ) = 4 Hence the required quadratic equation is x 2 = (sum of the roots) x + (Product of two roots) = 0 ⇒ x 2 - 3 √2x + 4 = 0.

Question 9

Ⅰ. 16x 2 + 20x + 6 = 0 Ⅱ. 10y 2 + 38y + 24 = 0

Accepted Answer

Question 10

Ⅰ. 18x 2 + 18x + 4 = 0 Ⅱ. 12y 2 + 29y + 14 =0

Accepted Answer

Question 11

Ⅰ. 4x + 7y = 209 Ⅱ. 12x – 14y = –38

Accepted Answer

Question 12

Ⅰ. 17x 2 + 48x = 9 Ⅱ. 13y 2 = 32y – 12

Accepted Answer

Question 13

Ⅰ. 8x 2 + 6x = 5 Ⅱ. 12y 2 – 22y + 8 = 0

Accepted Answer

Quadratic Equation Questions