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