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प्रश्न
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
उत्तर
Let the two parts be x and y.
From the given information,
= x + y = 20 ⇒ y = 20 – x
= 3x2 = (20 – x) + 10
= 3x2 = 30 – x
= 3x2 + x – 30 = 0
= 3x2 – 9x + 10x – 30 = 0
= x(3x + 10) – 3(3x + 10) = 0
= (3x + 10)(x – 3) = 0
= `x = 3, x = (-10)/3`
Since, x cannot be equal to `(-10)/3`, so, x = 3.
Thus, one part is 3 and the other part is 20 – 3 = 17.
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