1.

Find the values of k for which the given equation has real roots: (i)`kx^(2)-6x-2=0" "(ii)" "3x^(2)+2x+k=0" "(iii)" "2x^(2)+kx+2=0`

Answer» (i) The given equation is `kx^(2)-6x-2=0.`
`:." "D={(-6)^(2)-4xxkxx(-2)}=(36+8k).`
The given equation will have real roots if `Dge0.`
Now, `Dge0implies36+8kge0impliesimplieskge(-36)/(8)implieskge(-9)/(2).`
(ii) The given equation is `3x^(2)+2x+k=0.`
`:." "D=(2^(2)-4xx3xxk)=(4-12k).`
The given equation will have real roots if `Dge0.`
Now, `Dge0implies4-12kge0implies12kle4implieskle(1)/(3).`
(iii) The given equation is `2x^(2)+kx+2=0.`
`:." "D=(k^(2)-4xx2xx2)=(k^(2)-16).`
The given equation will have real roots if `Dge0.`
Now, `Dge0implies(k^(2)-16)ge0impliesk^(2)ge16implieskge4" or "kle-4.`


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