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Question
If `x=2/3` and x =−3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.
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Solution
The given equation is ax2 + 7x + b = 0.
Its roots are given as `x_1= −3` and `x_2=2/3`.
Now,
`=>x_1+x_2=-b/a`
`=>-3+2/3=(-(7))/a`
`=>(-9+2)/3=(-7)/a`
`=>(-7)/3=(-7)/a`
⇒ a = 3
Also
`=>x_1xxx_2=b/a`
`=>-3xx2/3=b/a`
`=>-2=b/3`
⇒ b = −6
Thus, the values of a and b are 3 and −6, respectively.
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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.
