Question

16x4-24x3+41x2-mx+16 be a perfect square then value of m is​

Answer 1

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Answer 2

$\large { \underline{ \bf \purple{Question:- }}}$

16x4-24x3+41x2-mx+16 be a perfect square then value of m is?

$\large { \underline{ \bf \pink{Solution:- }}}$

It is given that the polynomial 16x 4−24x 3 +41x 2−mx+16 to be a perfect square, therefore, we have

$16x4−24x3+41x2−mx+16=(ax2+bx+c)2$

$⇒16x4−24x3+41x2−mx+16=a2x4+b2x2+c2+2abx3+2cax2+2bcx$

$(∵(a+b+c)2=a2+b2+c2+2ab+2bc+2ca)$

Comparing the coefficient of x on both sides, we get:-

$a2=16⇒a=4$

$c2=16⇒c=4$

$2ab=−24$

$⟹2×4×b=−24$

$⟹8b=−24$

$⟹b = - \frac{24}{8}$

$⟹b = - 3$

$⟹2bc=−m$

$⟹2×−3×4=−m$

$⟹−24=−m$

$⟹m=24$

$\large { \underline{ \bf \red{Therefore }}}$

$m = 24 \: Is \: the \: Correct Answer$