Question

If (a? +b)x2 +2 (ab+bd)x +c+d2 = 0 has no real roots, then​

Answer 1

Given equation is

$\bf \: x2 ( {a}^{2} + {b}^{2} ) + 2x(ac + bd) + ( {c}^{2}+ {d}^{2}) = 0$

The equation will have no real roots only if Discriminant (D) < 0

$\bf⇒ b2 \: – \: 4ac < 0$

$\bf⇒ b2 \: – \: 4ac$

$\bf ⇒ [2(ac + bd)]2 \: – \: 4 ( {a}^{2}+ {b}^{2}) ( {c}^{2} + {d}^{2})$

$\bf ⇒ 4( {a \: c \: + b \: d}^{2}) \: – 4( {a}^{2} \: {c}^{2} + {a}^{2} \: {d}^{2} + {b}^{2} {c}^{2} + {b}^{2} {d}^{2})$

$\bf ⇒ 4( {a}^{2} {c}^{2} + {b}^{2} {d}^{2} + 2 \: abcd) – 4( {a}^{2} {c}^{2} + {a}^{2} {d}^{2} + {b}^{2} {c}^{2} + {b}^{2} {d}^{2})</p><p>$

$\bf ⇒ 4 \: {a}^{2} {c}^{2} + 4 \: {b}^{2} {d}^{2} + 8 \: abcd – 4 \: {a}^{2} {c}^{2} – 4 \: {a}^{2} {d}^{2} – \: 4 \: {b}^{2} {c}^{2} \: – 4 \: {b}^{2} {d}^{2}</p><p>$

$\bf ⇒ – 4 \: {a}^{2} {d}^{2} \: – \: 4 {b \:}^{2} {c}^{2} + 8 \: abcd$

$\bf – 4 [ {a}^{2} {d}^{2} + {b}^{2} {c}^{2} – 2 \: abcd]$

$\bf – 4[( {a \: d \: - \: b \: c)}^{2} ]$

For ad ≠ bc

$\bf D = – \: 4 × [Value \: of ( {a \: d \: - \: b \: c)}^{2}]$

∴ D always remain negative

So , D < 0

⇒ The given equation has no real roots.

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