The quadratic expression (2x+1)^{2} - px + q | vedclass

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Question

(C) Given the quadratic expression is $(2x+1)^{2} - px + q 
eq 0$ for all real $x$.Expanding the expression:$4x^{2} + 4x + 1 - px + q 
eq 0$$4x^{2}

Vedclass · Accepted Answer

$p^{2} - 8p - 16q < 0$

Vedclass · Answer

(C) Given the quadratic expression is $(2x+1)^{2} - px + q 
eq 0$ for all real $x$.Expanding the expression:$4x^{2} + 4x + 1 - px + q 
eq 0$$4x^{2} + (4-p)x + (1+q) 
eq 0$For a quadratic expression $ax^{2} + bx + c$ to be non-zero for all real $x$,its discriminant $D$ must be less than $0$.$D = b^{2} - 4ac Substituting the coefficients $a = 4$,$b = (4-p)$,and $c = (1+q)$:$(4-p)^{2} - 4(4)(1+q) $16 - 8p + p^{2} - 16 - 16q $p^{2} - 8p - 16q < 0$

Quadratic expression	The minimum value
i) $x^2 + 4x + 6$	a) $1$
ii) $x^2 - 2x + 5$	b) $2$
iii) $x^2 + 6x + 18$	c) $4$
iv) $x^2 - 4x + 5$	d) $9$

The quadratic expression $(2x+1)^{2} - px + q \neq 0$ for any real $x$,if

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i) $x^2 + 4x + 6$ a) $1$
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