Question

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Answer 1

$\sf\huge\bold{Answer:}$

$\fbox{Solution:1}$

Given :

x²-x-(k+5)= 0

Product of roots = -12

To find :

Value of k

Solution :

We know,

$\boxed{\tt\red{ product \: of \: roots = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}$

Constant term = -(k+5)

Coefficient of x² = 1

$\tt\ product \: of \: roots = \frac{ - (k + 5)}{1}$

$\tt\ product \: of \: roots \: = - k - 5$

• Inserting value of product of roots

$\tt\ - 12 = - k - 5$

$\tt\ - 12 + 5 = - k$

$\tt\ - 7 = - k$

$\tt\ \cancel{ - } \: 7 = \cancel{ - } \: k$

$\huge \boxed{\tt\purple {k = 7}}$

$\fbox{Solution:2}$

Given :

• px²-kx+p=0
• Have equal roots
• k > 0

To find :

• Value of p

Solution :

When quadratic equation have equal roots, discriminant is 0

» D = b²-4ac

» 0 = b²-4ac

General form of equation :

ax²+bx+c=0

Comparing given equation with general form

a = p, b = -k, c = p

• Inserting values of a, b, c in discriminant.

» 0 = (-k)²-4(p)(p)

» 0 = (-1)²k²-4p²

» 0 = k²-(2p)²

[ x²-y² = (x+y)(x-y) ]

» 0 = (k+2p)(k-2p)

» k+2p = 0, k-2p = 0

» 2p = -k, -2p = -k

» p = -k/2, p = -k/-2

» p = -k/2, p = k/2

$\huge \boxed{\tt\purple {p = \frac{-k}{\:\:2}, \frac{k}{2}}}$