Hello sir, Please help me out with my doubt

If a1, a2 and b1, b2 are roots of the equation
2x​2+3x+k1=0and x2+2x+k2=0 respectively, then the system of equation a1y+a2z=0 and b1y+b2z=0 has a non zero solution. What will be the value of k1:k2?

Dear Students,a1y+a2z=0     andb1y+b2z=0has a non-zero solution. It means that there slope are not equal or determinant A0where A=a1a2b1b20=a1×b2-a2×b10.........(i)Now, from 2x2+3x+k1=0a1, a2=-3±9-8k14and, x2+2x+k2=0b1, b2=-2±4-4k22=-1±1-k2Now from equation (i);(-3+9-8k14)(-1-1-k2)-(-3-9-8k14)(-1+1-k2)0it is in the form of (a+b)(c-d)-(a-b)(c+d)=2(bc-ad)So, 2[9-8k14×(-1)-{-34×1-k2}]0=12[31-k2-9-8k1]0=9-8k131-k2squaring both sides, we get;=9-8k19(1-k2)=9-8k19-9k2=8k19k21=k1k298Regards.

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