Question

6)
The equation x² -(p+4)X + 2p + 5 = 0 has equal roots the values of p will be.
a) 1
c) ± 2
d) -2
b)2​

Answer 1

AnswEr :

The value of p could be ± 2

Given,

$\sf \: {x}^{2} - (p + 4)x + 2p + 5 = 0$

The above equation has two equal roots

We have to find the value of P

On comparing with ax² + bx + c = 0,

• a = 1
• b = - (p + 4)
• c = 2p + 5

If a quadratic equation has equal roots,it's discriminant is equal to zero,

D = 0

» b² - 4ac = 0

» [-(p+4)]² = 4(1)(2p + 5)

» (p² + 8p + 16) = 8p + 20

» p² = 4

» p = ± 2

Answer 2

Answer:

${x}^{2} -( p + 4)x + 2p + 5 = 0 \\ = > {x}^{2} - px - 4x + 2p + 5 = 0 \\ = > {x}^{2} - (4 + p) x+(2p+ 5) = 0 \\ = > x = \frac{ ( 4 + p)( + - ) \sqrt{( 4 + p) {}^{2} - 4 \times 1 \times (2p + 5)} }{2 \times 1} \\ = > x = \frac{(4 + p)( + - ) \sqrt{(16 + 8p + {p}^{2} )- 4(2p + 5)} }{2} \\ = > x = \frac{(4 + p)( + - ) \sqrt{16 + 8p + {p }^{2} - 8p - 20 } }{2} \\ \\ = > x = \frac{(4 + p)( + - ) \sqrt{ {p }^{2} - 4 } }{2}$

$x1 = \frac{4 + p+ \sqrt{ {p}^{2} - 4} }{2} \\ x2 = \frac{4 + p - \sqrt{ {p}^{2} - 4 } }{2}$

the roots are equal

therefore,

$x1 = x2 \\ = > \frac{4 + p + \sqrt{ {p}^{2} - 4 } }{2} = \frac{4 + p - \sqrt{ {p}^{2} - 4} }{2} \\ = > 4 + p + \sqrt{p {}^{2} - 4} = 4 + p - \sqrt{p {}^{2} - 4} \\ = > \sqrt{ {p}^{2} - 4} = - \sqrt{ {p}^{2} - 4} \\ = > \sqrt{ {p}^{2} - 4 } + \sqrt{ {p}^{2} - 4} = 0 \\ = > 2 \sqrt{ {p}^{2} - 4 } = 0 \\ = > \sqrt{ {p}^{2} - 4} = 0 \\ = > {p}^{2} - 4 = 0 \\ = > {p}^{2} = 4 \\ = > p = + 2 \: \: \: \: | p = \: \: - 2$

so option B is correct (+-)2