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प्रश्न
Find whether the following equation have real roots. If real roots exist, find them.
`1/(2x - 3) + 1/(x - 5) = 1, x ≠ 3/2, 5`
उत्तर
Given equation is `1/(2x - 3) + 1/(x - 5) = 1, x ≠ 3/2, 5`
⇒ `(x - 5 + 2x - 3)/((2x - 5)(x - 5))` = 1
⇒ `(3x - 8)/(2x^2 - 5x - 10x + 25)` = 1
⇒ `(3x - 8)/(2x^2 - 15x + 25)` = 1
⇒ 3x – 8 = 2x2 – 15x + 25
⇒ 2x2 – 15x – 3x + 25 + 8 = 0
⇒ 2x2 – 18x + 33 = 0
On company with ax2 + bx + c = 0, we get
a = 2, b = – 18 and c = 33
∴ Discriminant, D = b2 – 4ac
= (–18)2 – 4 × 2(33)
= 324 – 264
= 60 > 0
Therefore, the equation 2x2 – 18x + 33 = 0 has two distinct real roots.
Roots, `x = (-b +- sqrt(D))/(2a)`
= `(-(-18) +- sqrt(60))/(2(2))`
= `(18 +- 2sqrt(15))/4`
= `(9 +- sqrt(15))/2`
= `(9 + sqrt(15))/2, (9 - sqrt(15))/2`
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संबंधित प्रश्न
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Without actually determining the roots comment upon the nature of the roots of each of the following equations:
x2 + 5x + 15 = 0.
Determine whether the given quadratic equations have equal roots and if so, find the roots:
x2 + 5x + 5 = 0
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(x – 1)(x + 2) + 2 = 0
Complete the following activity to determine the nature of the roots of the quadratic equation x2 + 2x – 9 = 0 :
Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
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