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
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Find 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
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2x2 + 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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Last updated at Aug. 3, 2021 by Teachoo
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