Reduce the expression y-x2-x-1x2-x-1 to the form ax2-bx-c and give condition for x to be real-

Y= x^2 x+1x^2+x+1 nbsp; ⇒ x^2(y 1)+x(y + 1)+(y 1) = 0 This is in the form of nbsp; ax^2 + bx +c =0. Condition for x to be real is D nbsp;≥ nbsp;0 ...

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Standard XIII

Mathematics

Discriminant of a Quadratic Equation

Reduce the ex...

Question

Reduce the expression $y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ to the form $a x^{2} + b x + c$ and give condition for $x$ to be real.

$(y + 1)^{2} - 4 (y - 1)^{2} \geq 0$

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$(y - 1)^{2} - 2 \geq 0$

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$(y - 1)^{2} + 4 (y + 1)^{2} \geq 0$

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$(y - 1)^{2} - 4 (y + 1)^{2} \geq 0$

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Solution

The correct option is A
$(y + 1)^{2} - 4 (y - 1)^{2} \geq 0$

$y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$

$\Rightarrow x^{2} (y - 1) + x (y + 1) + (y - 1) = 0$

This is in the form of $a x^{2} + b x + c = 0$ .

Condition for $x$ to be real is $D \geq 0$ .

$\Rightarrow b^{2} - 4 a c \geq 0$

$\Rightarrow (y + 1)^{2} - 4 (y - 1)^{2} \geq 0$

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Reduce the expression y=x2−x+1x2+x+1 to the form ax2+bx+c and give condition for x to be real.

The correct option is A (y+1)2−4(y−1)2≥0 y=x2−x+1x2+x+1 ⇒x2(y−1)+x(y+1)+(y−1)=0 This is in the form of ax2+bx+c=0. Condition for x to be real is D≥0 . ⇒b2−4ac≥0 ⇒(y+1)2−4(y−1)2≥0

Reduce the expression $y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ to the form $a x^{2} + b x + c$ and give condition for $x$ to be real.