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Given, (a^2+b^2)x^2 -2(ac + bd)x + (c^2 +d^2)=0 are equal.
We know that b^2-4ac = 0.
= (-2(ac + bd))^2 - 4(a^2+b^2)(c^2 + d^2) = 0
= 4(a^2c^2 + b^2d^2+2abcd) - 4(a^2c^2+a^2d^2+b^2c^2 +b^2d^2) = 0
On dividing with 4 into both sides, we get
= a^2c^2 + b^2d^2+2abcd-a^2c^2-a^2d^2-b^2c^2 -b^2d^2 = 0
= 2abcd-a^2d^2-b^2c^2 = 0
= a^2d^2+b^2c^2 - 2abcd = 0
= (ad-bc)^2 = 0
ad = bc
a/b = c/d.
LHS = RHS.
Hope this helps!
We know that b^2-4ac = 0.
= (-2(ac + bd))^2 - 4(a^2+b^2)(c^2 + d^2) = 0
= 4(a^2c^2 + b^2d^2+2abcd) - 4(a^2c^2+a^2d^2+b^2c^2 +b^2d^2) = 0
On dividing with 4 into both sides, we get
= a^2c^2 + b^2d^2+2abcd-a^2c^2-a^2d^2-b^2c^2 -b^2d^2 = 0
= 2abcd-a^2d^2-b^2c^2 = 0
= a^2d^2+b^2c^2 - 2abcd = 0
= (ad-bc)^2 = 0
ad = bc
a/b = c/d.
LHS = RHS.
Hope this helps!
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