Question

Find the values of k for which the equation x² + 5kx + 16 = 0 has no real roots .

class 10th mathematics.

Fast guys

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$


$\bf \pink{step \: by \: step \: explanation}$


The given equation is

→ x² + 5kx + 16 = 0


→ D = (5k)² - 4( 1 )( 16 ) = 25k² - 64


→ If the equation has no real roots , then D < 0.


$\bf{ \implies \: {25k}^{2} - 64 < 0 }$


$\bf{ \implies \: {k}^{2} < \frac{64}{25}}$

$\bf{ \implies \: k < \frac{8}{5} } \\ \\ \bf{ \implies \: k > \frac{ - 8}{5}}$


Hence, for no real roots ,

$\bf{ \frac{ - 8}{5} < k < \frac{8}{5}}$



$\huge \green{ \boxed{ \boxed{ \mathscr{THANKS}}}}$

Answer 2

$\underline{\mathfrak{\huge{Answer}}}$

${x^2 + 5kx + 16 = 0}$

From the above equation,

a = 1; b = 5k; c = 16

If the equation doesn't have equal roots,

${D = b^2 - 4ac < 0}$

Substituting the values,

${{5k}^2 - 4(16) <0}$

${25k^2 - 64 < 0}$

${25k^2 < 64}$

${k^2 < }$${\frac{64}{25}}$}

${k < }$${\frac{8}{5}}$

${k > }$${\frac{-8}{5}}$

Therefore ,

${\huge{\frac{-8}{5}}}$ > ${\huge{k}}$ > ${\huge{\frac{8}{5}}}$

Hence it is solved