This question is taken from quadratic equation chapter in which we have given an equation ax²+bx-6=0 and we have to find the value of a and b if the roots of the given equation are 3/4 and -2.

Kindly solve the above problem in detail

RS Aggarwal, Class 10, chapter 4A, question no 3(ii)

## The quadratic equation is ax²+bx−6=0 ……(i)

Given, x=3/4 and x=−2 are the roots of the quadratic equation.

Thus, putting x=3/4 in equation (i) we get,

a(3/4)²+b×3/4−6=0

⇒a×9/16+3/4b−6=0

⇒(9a+12b−96)/16=0

⇒9a+12b−96=0

⇒3a+4b−32=0 ….(ii)

Now, putting x=−2 in equation (i), we get

a(−2)²+b(−2)−6=0

⇒4a−2b−6=0

⇒2a−b−3=0 ….(iii)

Now, multiplying equation (iii) by 4, we get

⇒8a−4b−12=0 ….(iv)

Now, adding (ii) and (iv), we get

11a−44=0

⇒a=44/11

⇒a=4

Now, putting the value of p in equation (iii), we get

⇒2a−b−3=0

⇒2×4−b−3=0

⇒8−b−3=0

⇒−b=3−8

⇒−b=−5

⇒b=5