Question

find the values of p so that the equation (p-4) x-2 (p-4) x+4=0 has real roots​

Answer 1

GIVEN :-

• (p-4) x-2 (p-4) x+4=0 has real roots

TO FIND :-

• value of p

SOLUTION :-

WE KNOW THAT CONDITION FOR REAL ROOTS :-

$\implies \boxed{ \rm{ {b}^{2} - 4ac \geqslant \: 0}}$

hence,

$\implies \rm{(p - 4) ^{2} - (4)(4)(p - 4) \geqslant 0}$

now we know that IDENTITY :-

$\implies \boxed{\rm{(a - b) ^{2} = {a}^{2} + {b}^{2} } - 2ab}$

hence

$\implies \rm{(p {}^{2} + 16 - 8p) - (4)(4)(p - 4) \geqslant 0}$

$\implies \rm{(p {}^{2} + 16 - 8p) - (16)(p - 4) \geqslant 0}$

$\implies \rm{(p {}^{2} + 16 - 8p) - (16p - 64) \geqslant 0}$

$\implies \rm{(p {}^{2} + 16 - 8p) - 16p + 64\geqslant 0}$

$\implies \rm{p {}^{2} + 16 - 8p - 16p + 64\geqslant 0}$

$\implies \rm{p {}^{2} - 24p+ 80\geqslant 0}$

now by middle term break

$\implies \rm{p {}^{2} -( 20 + 4)p+ 80\geqslant 0}$

$\implies \rm{p {}^{2} -20p - 4p+ 80\geqslant 0}$

$\implies \rm{p \: (p -20) - 4 \: (p - 20)\geqslant 0}$

hence , (1)

$\implies \rm{ \: (p -20) (p - 4)\geqslant 0}$

$\implies \rm{ \: (p -20) \geqslant 0}$

$\implies \boxed{ \rm{ \: p \geqslant 20}}$

now again (2)

$\implies \rm{ \: (p -20) (p - 4)\geqslant 0}$

$\implies \rm{ \: (p -4) \geqslant 0}$

$\implies \boxed{ \rm{ \: p \geqslant 4}}$

HENCE VALUES OF P ARE :-

$\implies \boxed{\boxed{ \rm{ \: p \geqslant 4 \: and \:\: p \geqslant20}}}$

OTHER INFORMATION :-

Discriminant :

• For a quadratic equation of the form ax²+bx+c=0, the expression b²−4ac is called the discriminant, (denoted by D), of the quadratic equation.

• The discriminant determines the nature of roots of the quadratic equation based on the coefficients of the quadratic equation.

nature of roots :

Based on the value of the discriminant, D=b2−4ac, the roots of a quadratic equation can be of three types.

• Case 1: If D>0, the equation has two distinct real roots.

• Case 2: If D=0, the equation has two equal real roots.

• Case 3: If D<0, the equation has no real roots.