Question

Solve it ^^

​

Answer 1

Answer:

Ans is in the pic.

Step-by-step explanation:

By the help of my cousin because he is in 12th class and I'm in 8th class.

Answer 2

The Given equation is (a² + b²)x² + 2(ac + bd)x + (c² + d²)

$\begin{array}{l l l} \\ \large\sf \bf \fbox \red{Discriminant = b²-4ac} \\ \\ \mapsto \sf{D=2(ac+bd) {}^{2} −4(a {}^{2} +b {}^{2} )(c {}^{2} +d {}^{2} )} \\ \\\mapsto \sf{D=4(ac+bd) {}^{2} −4(a {}^{2} +b {}^{2} )(c {}^{2} +d {}^{2} )} \\\\\mapsto\sf{D=4[(ac+bd) {}^{2} −(a {}^{2} +b {}^{2} )(c {}^{2} +d {}^{2} )] }\\\\ \mapsto\sf{ D=4[ \cancel {a {}^{2} c {}^{2} }+ \cancel {b {}^{2} d {}^{2}} +2acbd \cancel {−a {}^{2} c {}^{2}} −a {}^{2} d {}^{2} −b {}^{2} c {}^{2} \cancel{−b {}^{2} d {}^{2}} ]} \\\\\mapsto \sf{D=4[2acbd−a {}^{2} d {}^{2} −b {}^{2} c {}^{2} ] }\\\\\mapsto\sf{D=−4[a {}^{2} d {}^{2} +b {}^{2} c {}^{2} −2adbc]} \\\\ \mapsto\sf \bf \fbox{D \: = \: −4(ad−bc)²} \\ \\\sf \bf{We \: \: have,  ad \neq \: bc}\\\\ \rightarrow\sf{ad−bc=0} \\\\ \rightarrow\sf{(ad−bc)² \: >0} \\\\\rightarrow\sf{−4(ad−bc)²<0 }\\\\\rightarrow\sf \bf \fbox{D< \: 0 }\\ \\ \bf \sf \fbox \blue{Hence, \: the \: given \: equation \: has \: no \: real \: roots.} \end{array}</p><p></p><p>$