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# If ( a 2 + B 2 ) X 2 + 2 ( a B + B D ) X + C 2 + D 2 = 0 Has No Real Roots, Then - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

If $\left( a^2 + b^2 \right) x^2 + 2\left( ab + bd \right)x + c^2 + d^2 = 0$ has no real roots, then

• ab = bc

• ab = cd

• ac = bd

#### Solution

The given quadric equation is  $\left( a^2 + b^2 \right) x^2 + 2\left( ab + bd \right)x + c^2 + d^2 = 0$ , and roots are equal.

Here, a = (a^2 + b^2 ),b = 2 (ab + bd) and , c = c^2 + d^2

As we know that D = b^2 - 4ac

Putting the value of  a = (a^2 + b^2 ),b = 2 (ab + bd) and , c = c^2 + d^2

={2 (ab + bd)}^2 - 4 xx (a^2 _b^2) xx (c^2 + d^2)

 = 4a^2b^2 + 4b^2d^2 + 8ab^2d - 4(a^2c^2 + a^2 d^2 +b^2c^2 + b^2 d^2)

=4a^2b^2 + 4b^2d^2 + 8ab^2d - 4a^2c^2 - 4a^2d^2 - 4b^2 c^2 - 4b^2d^2

= 4a^2b^2 + 8ab^2 d - 4a^2c^2 - 4a^2d^2 - 4b^2c^2

= 4 (a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2)

The given equation will have no real roots, if D < 0

4 (a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2) < 0

a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2) < 0

Is there an error in this question or solution?

#### APPEARS IN

Solution If ( a 2 + B 2 ) X 2 + 2 ( a B + B D ) X + C 2 + D 2 = 0 Has No Real Roots, Then Concept: Solutions of Quadratic Equations by Factorization.
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