Question

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$\huge{ ➢QUESTION }$
Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2) = 0 has no real root if ad ≠ bc​

Answer 1

Answer:

Given Question :

Prove that $\mathrm{x^2(a^2+b^2)+2x(ac+bd)+(c^2+d^2) = 0}$ has no real root if $\mathrm{ad \neq bc}$

Solution :

Given quadratic equation,

$\longmapsto \mathrm{x^2(a^2+b^2)+2x(ac+bd)+(c^2+d^2) = 0}$

If it is the form of Ax² + Bx + C = 0, then

$\mathrm{A = a^2+b^2\ ; \ B = 2(ac+bd)\ ; \ C = c^2+d^2}$

We know that,

$\boxed{\begin{array}{cc} \mathrm{Discriminant = \Delta = B^2-4AC}\\\\\underline{\text{Conditions are :}}\\\implies \Delta > 0\ ;\ \text{Roots are real and distinct}\\\implies \Delta = 0\ ;\ \text{Roots are real and equal}\\\implies \Delta < 0\ ;\ \text{Roots are not real} \end{array}}$

To know the given equation satisfies any of the three conditions, let us find the value of the discriminant.

$\implies \mathrm{\Delta = \big(2(ac+bd)\big)^2-4(a^2+b^2)(c^2+d^2)}$

We know that,

$\boxed{\mathrm{(x+y)^2 = x^2+2xy+y^2}}$

So,

$\implies \mathrm{\Delta = 4(\cancel{a^2c^2} + 2abcd+\cancel{b^2d^2})-4(\cancel{a^2c^2} + a^2d^2+b^2c^2+\cancel{b^2d^2})}$

$\implies \mathrm{\Delta = 8abcd-4(a^2d^2+b^2c^2)}$

Taking '-4' common,

$\implies \mathrm{\Delta = -4(-2abcd+(a^2d^2+b^2c^2))}$

$\implies \mathrm{\Delta = -4(a^2d^2-2abcd+b^2c^2)}$

Bracket is of the form $\mathrm{x^2-2xy+y^2}$. So, it can be written as,

$\implies \mathrm{\Delta = -4(ad-bc)^2}$

As, $\boxed{\mathrm{x^2-2xy+y^2 = (x-y)^2}}$

Thus,

$\implies \mathrm{\Delta = -4(ad-bc)^2}$

This can't get cancelled as $\mathrm{ad \neq bc}$

We also know that, square is always > 0. So the bracket and it square always keeps the number >0. The thing which matters here is '-4' which is out of the bracket.

As, it is negative, we can say that,

$\implies \mathrm{\Delta = -4(ad-bc)^2 < 0}$

$\implies \mathrm{\Delta < 0}$

So, this follows the 3rd condition of the box and proves that it has no real roots

Thus, $\mathrm{x^2(a^2+b^2)+2x(ac+bd)+(c^2+d^2) = 0}$ has no real root if $\mathrm{ad \neq bc}$

${\large{\textsf{\underline{\underline{Hence Proved}}!!}}}$

Hope it helps!

Answer 2

Answer:

They were either one or two storeys high, with rooms built around a courtyard.

Most houses had a separate bathing area.

Most houses had a separate bathing area.Some houses had wells to supply water.

Every house had dustbins so that no garbage was littered.

Houses had drain pipes which were connected to the main drain.

Step-by-step explanation:

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