1.

Show that the equation `9x^(2)+7x-2=0` has roots and solve it.

Answer» The given equation is `9x^(2)+7x-2=0.`
Comparing it with `ax^(2)+bx+c=0`, we get
`a=9,b=7" and "c=-2`
`:." "D=(b^(2)-4ac)=(7^(2)-4xx9xx(-2)]=121gt0.`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt(121)=11.`
`:." "alpha=(-b+sqrt(D))/(2a)=((-7+11))/(2xx9)=(4)/(18)=(2)/(9),`
`beta=(-b-sqrt(D))/(2a)=((-7-11))/(2xx9)=(-18)/(18)=-1.`
Hence, the required roots are `(2)/(9)` and -1.


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