Question

If One root of the quadratic equation (3x-2)(2x+1)=0 is then

the other root is​

Answer 1

Answer:

${tex}9</h2><h2></h2><h2>3x2+ax−2=0</h2><h2></h2><h2>Since, one root is 1, then x=1</h2><h2></h2><h2>⇒  3(1)2+a(1)−2=0</h2><h2></h2><h2>⇒  3+a−2=0</h2><h2></h2><h2>⇒  a+1=0</h2><h2></h2><h2>⇒  a=−1</h2><h2></h2><h2>⇒  Now, it is given that ax2+6ax−b=0 has equal roots.</h2><h2></h2><h2>∴  b2−4ac=0</h2><h2></h2><h2>⇒  (6a)2−4(a)(−b)=0</h2><h2></h2><h2>⇒  36a2+4ab=0</h2><h2></h2><h2>⇒  36(−1)2+4(−1)b=0                 [ Substituting a=−1 ]</h2><h2></h2><h2>⇒  36−4b=0</h2><h2></h2><h2>⇒  4b=36</h2><h2></h2><h2>∴  b=9</h2><h2></h2><h2>{\tex}$

hope it will help u

Answer 2

Answer:

2/3 , -1/2

Step-by-step explanation:

For finding roots of quadratic equation, put each factor equal to 0

that is,

3x - 2 = 0

and

2x + 1 = 0

3x-2 = 0

x = 2/3

2x+1 = 0

x = -1/2

Roots of quadratic equation, (3x-2)(2x+1) = 0 are 2/3 and -1/2