Pls help with question 5

Dear Student,

x^{2} + b_{1} x + c_{1} = 0 For this equation to have real roots {b_{1}}^{2} - 4 c_{1} \geq 0 \Rightarrow {b_{1}}^{2} \geq 4 c_{1}; inequality (i) x^{2} + b_{2} x + c_{2} = 0 For this equation to have real roots {b_{2}}^{2} - 4 c_{2} \geq 0 \Rightarrow {b_{2}}^{2} \geq 4 c_{2}; inequality (ii) The relation should be b_{1} b_{2} = 2 (c_{1} + c_{2}) \Rightarrow {b_{1}}^{2} {b_{2}}^{2} = 4 {(c_{1} + c_{2})}^{2} Now we know that {(c_{1} + c_{2})}^{2} = {(c_{1} - c_{2})}^{2} + 4 c_{1} c_{2} \Rightarrow {b_{1}}^{2} {b_{2}}^{2} = 4 \{{(c_{1} - c_{2})}^{2} + 4 c_{1} c_{2}\} \Rightarrow {b_{1}}^{2} {b_{2}}^{2} = 4 {(c_{1} - c_{2})}^{2} + 16 c_{1} c_{2} \Rightarrow {b_{1}}^{2} {b_{2}}^{2} - 16 c_{1} c_{2} = 4 {(c_{1} - c_{2})}^{2} As R . H . S . \geq 0 \Rightarrow {b_{1}}^{2} {b_{2}}^{2} - 16 c_{1} c_{2} \geq 0 \Rightarrow {b_{1}}^{2} {b_{2}}^{2} \geq 16 c_{1} c_{2} If you observe inequality (i) \times inequality (ii) will also yield same result . \Rightarrow Atleast one of the equations will have real root . \Rightarrow Option (a) is correct .

Regards