Text Solution Solution : (i) `2x^2 – 7x + 3 = 0` <br> `⇒ 2x^2 – 7x = – 3` <br> On dividing by 2 on both sides, we get <br> `⇒ x^2 -(7x)/2 = -3/2` <br> `⇒ x^2 -2 × x ×7/4 = -3/2` <br> On adding` (7/4)^2` to both sides of the equation, we get <br> `⇒ (x)^2-2×x×7/4 +(7/4)^2 = (7/4)^2-3/2` <br> `⇒ (x-7/4)^2 = (49/16) – (3/2)` <br> `⇒(x-7/4)^2 = 25/16` <br> `⇒(x-7/4)^2 = ±5/4` <br> `⇒ x = 7/4 ± 5/4` <br> `⇒ x = 7/4 + 5/4 `or `x = 7/4 – 5/4` <br> `⇒ x = 12/4` or `x = 2/4` <br> `⇒ x = 3` or `x = 1/2` <br> ii) `2x^2 + x – 4 = 0` <br> `⇒ 2x^2 + x = 4` <br> On dividing both sides of the equation by 2, we get <br> `⇒ x^2 +x/2 = 2` <br> Now on adding (1/4)2 to both sides of the equation, we get, <br> ⇒ `(x)^2 + 2 × x × 1/4 + (1/4)^2 = 2 + (1/4)^2` <br> ⇒ `(x + 1/4)^2 = 33/16` <br> ⇒ `x + 1/4 = ± sqrt(33/4)` <br> ⇒ `x = ± sqrt(33/4) – 1/4` <br> ⇒` x = (± sqrt(33)-1)/4` <br> Hence, either `x = (sqrt(33)-1)/4` or `x = (-sqrt(33)-1)/4` <br> iii) Given, ` 4x^2 + 4sqrt3x + 3 = 0` <br> Converting the equation into `a^2+2ab+b^2` form, we get, <br> `⇒ (2x)^2 + 2 × 2x × √3 + (sqrt3)^2 = 0` <br> `⇒ (2x + sqrt3)^2 = 0` <br> `⇒ (2x + sqrt3) = 0` and `(2x + sqrt3) = 0` <br> Hence, either `x = -sqrt3/2` or `x = -sqrt3/2`. <br> iv) <br> `2x^2 + x + 4 = 0` <br> `⇒ 2x^2 + x = -4` <br> On dividing both sides of the equation by 2, we get <br> `⇒ x^2 + 1/2x = -2` <br> `⇒ x^2 + 2 × x × 1/4 = -2` <br> By adding `(1/4)^2` to both sides of the equation, we get <br> `⇒ (x)^2 + 2 × x × 1/4 + (1/4)^2 = (1/4)^2 – 2` <br> `⇒ (x + 1/4)^2 = 1/16 – 2` <br> `⇒ (x + 1/4)^2= -31/16` <br> As we know, the square of numbers cannot be negative. <br> Hence, there is no real root for the given equation, `2x^2 + x + 4 = 0`. Find the roots of the following quadratic equations, if they exist, by the method of completing the square 2x2 + x – 4 = 0 2x2 + x – 4 = 0 ⇒ 2x2 + x = 4 On dividing both sides of the equation, we get `⇒ x^2 + x/2 = 2` On adding (1/4)2 to both sides of the equation, we get `⇒ (x)^2 + 2 × x × 1/4 + (1/4)^2 = 2 + (1/4)^2` `⇒ (x + 1/4)^2 = 33/16` `⇒ x + 1/4 = ± sqrt33/4` `⇒ x = ± sqrt33/4 - 1/4` `⇒ x = ± sqrt33-1/4` `⇒ x = (sqrt33-1)/4 ` Concept: Solutions of Quadratic Equations by Completing the Square Is there an error in this question or solution? Page 2Find the roots of the following quadratic equations, if they exist, by the method of completing the square `4x^2 + 4sqrt3x + 3 = 0` `4x^2 + 4sqrt3x + 3 = 0` `⇒ (2x)^2 + 2 × 2x × sqrt3 + (sqrt3)^2 = 0` `⇒ (2x + sqrt3)^2 = 0` `⇒ (2x + sqrt3) = 0 ` `⇒ x = (-sqrt3)/2 ` Concept: Solutions of Quadratic Equations by Completing the Square Is there an error in this question or solution? Page 32x2 + x + 4 = 0 ⇒ 2x2 + x = -4 On dividing both sides of the equation, we get `⇒ x^2 + 1/(2x) = 2` `⇒ x^2 + 2 × x × 1/4 = -2` On adding (1/4)2 to both sides of the equation, we get `⇒ (x)^2 + 2 × x × 1/4 + (1/4)^2 = (1/4)^2 - 2 ` `⇒ (x + 1/4)^2 = 1/16 - 2` `⇒ (x + 1/4)^2 = -31/16` However, the square of number cannot be negative. Therefore, there is no real root for the given equation
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