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Question
Which of the following equations has no real roots?
Options
`x^2 - 4x + 3sqrt(2) = 0`
`x^2 + 4x - 3sqrt(2) = 0`
`x^2 - 4x - 3sqrt(2) = 0`
`3x^2 + 4sqrt(3)x + 4 = 0`
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Solution
`bb(x^2 - 4x + 3sqrt(2) = 0)`
Explanation:
(A) The given equation is `x^2 - 4x + 3sqrt(2)` = 0
On comparing with ax2 + bx + c = 0, we get
a = 1, b = – 4 and c = `3sqrt(2)`
The discriminant of `x^2 - 4x + 3sqrt(2)` = 0 is
D = b2 – 4ac
= `(-4)^2 - 4(1)(3sqrt(2))`
= `16 - 12sqrt(2)`
= 16 – 12 × (1.41)
= 16 – 16.92
= – 0.92
⇒ b2 – 4ac < 0
(B) The given equation is `x^2 + 4x - 3sqrt(2)` = 0
On comparing the equation with ax2 + bx + c = 0, we get
a = 1, b = 4 and c = `-3sqrt(2)`
Then, D = b2 – 4ac
= `(-4)^2 - 4(1)(-3sqrt(2))`
= `16 + 12sqrt(2) > 0`
Hence, the equation has real roots.
(C) Given equation is `x^2 - 4x - 3sqrt(2)` = 0
On comparing the equation with ax2 + bx + c = 0, we get
a = 1, b = – 4 and c = `-3sqrt(2)`
Then, D = b2 – 4ac
= `(-4)^2 - 4(1)(-3sqrt(2))`
= `16 + 12sqrt(2) > 0`
Hence, the equation has real roots.
(D) Given equation is `3x^2 + 4sqrt(3)x + 4` = 0
On comparing the equation with ax2 + bx + c = 0, we get
a = 3, b = `4sqrt(3)` and c = 4
Then, D = b2 – 4ac
= `(4sqrt(3))^2 - 4(3)(4)`
= 48 – 48
= 0
Hence, the equation has real roots.
Hence, `x^2 - 4x + 3sqrt(2)` = 0 has no real roots.
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