Question

If the zeroes of the quadratic polynomial x²+(a+1)x+ bare 2 and -3, then find the value of a and b.​

Answer 1

$\huge\sf \red{Solution:-}$

Let 2 zeroes be a and b of polynomial,

x² + px + q = 0

sum of roots = a + b = -p/1 = -p products of roots = ab = q/1 = q

(a + b)² = a² + b² + 2ab = a² + b² = p² - 2q

(a-b)² = a² + b² -2ab = p² - 2q -2q = p² -4q

now the question asks for new quadratic eq whose roots are (a+b) ² and (a-b)² 2

so sum of new roots are = (a+b)² + (a-b)²

so sum of new roots are = (a+b)² + (a-b)² =p² + p² -4q = 2p² - 4q

and product of roots = (a+b)²(a-b)² = 4 (p²)² (p²-4q)² = pª (p² +16q² + 8p²q)

hence new quadratic eq gonna be = x²-x(sum of roots) + (products of roots)

x²-x(2p² - 4q) + pª(p4 + 16q² +8p²q) = 0

Answer 2

Given :-

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

To Find :-

• Value of a and b

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

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

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

• The value of a is 0 and b is -6

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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