Question

100 points //
If the zeroes of the polynomial x³ - 3x² + x +1 are a-b, a, a+b. Find a & b.

Class 10th

Answer 1

If the zeroes of the polynomial x³ - 3x² + x +1 are a-b, a, a+b. Find a & b.

Good question,

Here is your perfect answer!

Since sum of zeroes of cubic polynomial = - (coefficient of x²) /coefficient of x³

=) (a-b) + a + (a+b) = - (-3)/1

=) 3a = 3

=) a = 1,

Since product of zeroes of cubic polynomial = - (constant term) /coefficient of x³

=) (a-b) a(a+b) = - 1/1

=) a(a²-b²) = - 1

=) 1(1² - b²) = - 1

=) 1 - b² = - 1

=) 1 + 1 = b²

=) 2 = b²

=) b = ±$\sqrt{2}$

Hence a = 1, b = ±$\sqrt{2}$

Answer 2

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\bf{QUESTION}$
If the zeroes of the polynomial x³ - 3x² + x +1 are a-b, a, a+b. Find a & b.

$\bf{step \: by \: step \: explanation}$

$P(x)= x^3-3x^2+x+1$

$\bf{ \alpha + \beta + \gamma = \frac{ - b}{a}}$

$\bf{LET \: \alpha = a - b \: }$

$\bf{ \beta = a}$

$\bf{ \gamma = a + b}$

$\bf{a-b+a+a+b = \frac{ - b}{a}}$

$\huge \bf{3a = \frac{ - ( - 3)}{1}}$

$\huge \bf {3a = \frac{3}{1}}$

$\huge \bf{a = \frac{3}{3}}$

$\huge \bf \orange{a = 1.}$

NOW

$\sf{ \implies \alpha \times \beta \times \gamma = \frac{ - d}{a}}$

$\sf{ \implies a - b \times a \times a + b = - 1}$

$\bf{put \: the \: value \: of \: a}$

$\sf{ \implies 1 \times - b \times 1 \times 1 \times b = - 1}$

$\sf{ \implies - b ^{2} + 1 = - 1}$

$\sf{ \implies - b ^{2} = - 1 - 1}$

$\sf{ \implies - b ^{2} = - 2 }$

$\sf{ \implies \cancel{ - }{b}^{2} = \cancel{ - }2}$

$\huge \sf{ \therefore b = \±\sqrt{2}}$

$\huge \bf \pink{ \mid{THANKS} \mid}$