Question

Write the quadratic equation whose roots are 3 and -1, and whose leading coefficient is 3

Answer 1

Concept:

$\text{We know that the two roots of a quadratic equation }ax^2+bx+c=0\text{ are:}\\\\x_1=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\text{ and }x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}.$'

$\text{When we add them,}\\x_1+x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}+\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\\text{or, }x_1+x_2=\dfrac{-b+\sqrt{b^2-4ac}-b-\sqrt{b^2-4ac}}{2a}=\dfrac{-2b}{2a}\\\text{or, }x_1+x_2=-\dfrac{b}{a}$

$\text{When we multiply them, }\\x_1.x_2=\bigg(\dfrac{-b+\sqrt{b^2-4ac}}{2a}\bigg)\bigg(\dfrac{-b-\sqrt{b^2-4ac}}{2a}\bigg)\\\\\text{or, }x_1.x_2=\dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2}=\dfrac{b^2-b^2+4ac}{4a^2}=\dfrac{4ac}{4a^2}\\\therefore\ x_1.x_2=\dfrac{c}{a}$

$\text{Hence, if }p\text{ and }q\text{ be the roots of the quadratic equation }ax^2+bx+c=0,\\p+q=-\dfrac{b}{a}\\\\\text{and }pq=\dfrac{c}{a}$

Answer:

$3x^2-6x-9=0$

Step-by-step Explanation:

$\text{Solution:}\\\text{Let the required quadratic equation be }3x^2+bx+c=0\\\text{Since 3 and -1 are the roots of this equation, }\\3+(-1)=-\dfrac{b}{3}\\\text{or, }2=-\dfrac{b}{3}\\\\\text{or, }b=-6$

$\text{And }3(-1)=\dfrac{c}{a}=\dfrac{c}{3}\\\\\text{or, }-3=\dfrac{c}{3}\\\\\text{or, }c=-9$

$\text{Our quadratic equation was: }\\3x^2+bx+c=0\\\text{Now we put the values of }b\text{ and }c,\\\therefore3x^2-6x-9=0\text{ is the required quadratic equation.}$