Question

00:18:16 III 16. What is the degree of the polynomial ax2 + x + b, if the zeroes of the quadratic polynomial+(a+1)x+ bare 2 and -3?​

Answer 1

Given :-

• Zeroes of the polynomial p(x) = x² + (a + 1)x + b are 2 and -3

To Find :-

• What is the degree of the polynomial?

Formula to be used :-

$\boxed{\pink{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\boxed{\pink{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

Solution :-

$\qquad$ ☀️We are given, Zeroes of the polynomial p(x) = x² + (a + 1)x + b are 2 and -3.

$\qquad\small\underline{\pmb{\sf \:According \: to \: the \: question :-}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{α + β = \dfrac{-b}{a}}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ -3 + 2 = \dfrac{-(a + 1)}{1}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ -1 = -a - 1}}$

$\qquad\leadsto\quad \pmb {\mathfrak{a = -1 + 1}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{a = 0}}}$

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$\pink{\qquad\leadsto\quad \pmb {\mathfrak{αβ = \dfrac{c}{a}}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{-3 × 2 = \dfrac{ b}{1}}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{b = -6}}}$

• Therefore, degree of polynomial x-6 will be 1.

Verification :-

$\green{\qquad\leadsto\quad \pmb {\mathfrak{ x² + (0 + 1)x + (-6)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ x² + x - 6 = 0}}$

$\qquad\leadsto\quad \pmb {\mathfrak{x² - 2x + 3x - 6 }}$

$\qquad\leadsto\quad \pmb {\mathfrak{x(x - 2) + 3(x - 2)}}$

$\qquad\leadsto\quad \pmb {\mathfrak{(x + 3)(x - 2)}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ x + 3 = 0 \: and\: x - 2 = 0}}$

$\green{\qquad\leadsto\quad \pmb {\mathfrak{ x = -3\: and \: x = 2}}}$

• Hence, Verified..!!

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$\qquad \qquad \:\bigstar \:\:\underline {\pmb{ \: Know \: More \:\::-}}\:$

• The quadratic formula is _

$\boxed{\pmb {\mathfrak { x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a} }}}$

• It can be written as :-

$\qquad$ $\pmb {\mathfrak{ \alpha = \dfrac{-b + \sqrt{b^2 - 4ac }}{ 2a }} }$

$\qquad$ $\pmb {\mathfrak{\beta = \dfrac{ - b - \sqrt{ b^2 - 4ac }}{ 2a } }}$

Where:-

• α , β are the roots of the quadratic equation . b² - 4ac is a discriminate .

The conditions are as follows :-

$\qquad$ ☀️If D = 0

• The roots are equal and real .

$\qquad$ ☀️If D > 0

• The roots are unequal and rational ( if it is a perfect square )

$\qquad$ ☀️If D > 0

• The roots are distinct and irrational ( if it is not a perfect square )

$\qquad$ ☀️If D < 0

• The roots are unequal and imaginary .

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