If a12-b12-c12-0 then the value of a-b-c2 is -

The correct option is A 2√(ab) √(a) +√(b) √(c)=0 √(a) +√(b)=√(c) On squaring, we get (√(a) +√(b))^2=(√(c))^2 a+b+2√(ab)=c So, a+b ...

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Vector Triple Product

If a12+b12-...

Question

If $a^{\frac{1}{2}} + b^{\frac{1}{2}} - c^{\frac{1}{2}} = 0$ then the value of $(a + b - c)^{2}$ is :

- 2 \sqrt{a b}

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2 b c

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4 a b

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4 a c

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Solution

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a^{\frac{1}{2}} + b^{\frac{1}{2}} - c^{\frac{1}{2}} = 0

(a + b - c^{2})

0 < a, b, c < 1

a + b + c = 2

\frac{a}{1 - a} \times \frac{b}{1 - b} \times \frac{c}{1 - c}

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

a_{3} x^{2} + b_{3} x + c_{3} = 0

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

P (x) = a_{1} x^{2} + 2 b_{1} x + c_{1}

Q (x) = a_{12} x^{2} + 2 b_{2} x + c_{2}

a_{1} \neq 0, a_{2} \neq 0

P (x)

Q (x)

\forall \times \in R g (x) = a_{1} a_{2} x^{2} + b_{1} b_{2} x + c_{1} c_{2}

Unique point is obtained for the pair of equations $a_{1}$ x + $b_{1}$ y + $c_{1}$ = 0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ = 0 if __________ .

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