Question

If a + b = -3 and ab = -5/2

then find the quadratic equation whose roots
are a and b.

Answer 1

Given :

• Sum of zeroes ( a + b ) = -3

• Product of Zeroes ( ab) = -5/2

To find

• The quadratic equation .

Solution :

As we know that :

• Quadratic equation

=> x² - ( sum of Zeroes )x + product of zeroes =0

Put the given values :

=> x² - (-3)x + ( -5/2) = 0

=> x² + 3x -5/2 = 0

=> (2x² + 6x -5) / 2 = 0

=> 2x² + 6x -5 = 0

Hence , the quadratic equation is 2x² + 6x -5 .

Answer 2

Answer:

• Sum of Zeroes ( α + β ) = - 3
• Product of Zeroes ( αβ ) = - 5/2
• Quadratic Polynomial = ?

⠀

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Polynomial=x^2-(Sum\:of\:Zeroes)x+Product\:of\:Zeroes\\\\\\:\implies\sf Polynomial=x^2 -(\alpha + \beta)x + ( \alpha \beta)\\\\\\:\implies\sf Polynomial=x^2 - ( - 3)x +\bigg(-\dfrac{5}{2}\bigg)\\\\\\:\implies\sf Polynomial=x^2 + 3x - \dfrac{5}{2}\\\\{\scriptsize\qquad\bf{\dag}\:\:\frak{multiplying \:\:each \:\:by\:\:2}}\\\\:\implies\sf Polynomial=(2 \times x^2) + (2 \times 3x) - \bigg(2 \times \dfrac{5}{2}\bigg)\\\\\\:\implies\underline{\boxed{\sf Polynomial = 2x^2 + 6x - 5}}$

⠀

$\therefore\:\underline{\textsf{Required quadratic equation is \textbf{2x ^\text2 + 6x - 5}}}.$

$\rule{180}{1.5}$

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\sf with zeroes \alpha\:\sf and\:\beta \\\\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta= \dfrac{ - \:b}{a}\:\:\bigg\lgroup\bf Sum\:of\:Zeroes\bigg\rgroup \\\\\\{\textcircled{\footnotesize2}} \: \:\alpha \beta= \sf\dfrac{c}{a}\:\:\bigg\lgroup\bf Product\:of\:Zeroes\bigg\rgroup\end{minipage}}$