The roots of the equation $x^2-3x-m(m+3)=0$ , where $m$ is a constant are

Question

a) $m,m+3$

b) $-m, m+3$

c) $m, -(m+3)$

d) $-m , -(m +3)$

these are the options can someone please help me out on this. this a question relating to quadratic equations studied in class 10

Answer 1

$$x^2-3x-m(m+3)=x^2+mx-(m+3)x-m(m+3)=$$ $$=x(x+m)-(m+3)(x+m)=(x+m)(x-m-3).$$ Can you kill it now?

Answer 2

Hint. Note that the monic quadratic equation with roots $a$ and $b$ can be written as $$(x-a)(x-b)=x^2-(a+b)x+ab.$$

Answer 3

Well a trick is to simply notice that the roots of the above quadratic equation satisfy the quadratic equation,so you can plug in the roots into the equation and check if they equate to zero!

Other way is if you are unaware of the relationship between the roots and the coefficients of the quadratic equation is to know it now,as this will help you very much ahead!

that is if $\alpha ,\beta$ are the roots of the quadratic equation $ax^2 + bx +c =0$ then $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.

so in your question $a = 1,b = -3, c = -m(m-3)$ so that $\alpha + \beta = 3$ and $\alpha \beta = -m(m-3)$ , you can check the options or can come to conclusion by solving the two equations to get $\alpha = m , \beta = -(m-3)$

Or you can use the formula of the roots of the quadratic equation!