Question

2. What is the value of k, if the roots of kx² +24x+16=0 are real and equal?
(A) 9
(B) -9
(C) 25
(D) – 25​

Answer 1

Step-by-step explanation:

option B is correct

-9 is the correct ans

Answer 2

Answer:

$p(x) = k {x}^{2} + 24x + 16 \\ given \: \: the \: roots are \: real \: and \: equal.so \: \: discriminant = 0 \\ \\ in \: order \: to \: find \: the \: value \: of \: k \: we \: have \: to \: find \: d \: first. \\ d = {b}^{2} - 4ac \\ a = k \: \: b = 24 \: \: c = 16 \\ \\ {b}^{2} - 4ac = ( {24}^{2} ) - 4 \times k \times 16 \\ = > 576 - 64k \\ \\ as \: given \: roots \: are \: real \: and \: equal \: so \: 576 - 64k = 0 \\ = > 64k = 576 \\ = > k = \frac{576}{64} \\ = > k = 9$

So, the required value of k = 9.