Question

if p and q are the zeros of 2x² - 7x + 3. find p²+ Q²​

Answer 1

Answer:

$\implies{ p^{2} + {q}^{2} = \frac{37}{4}}$

Step-by-step explanation:

Given:

${\boxed{\bold{\mathsf{2x^{2} - 7x + 3}}}}$

To find: p² + q²

Apply middle term splitting

$\boxed{\bold{\mathsf{2x^{2} - 7x + 3}}}$

$\implies{\bold{\mathsf{2x^{2} -6x -x +3 }}}$

$\implies$x(2 x (x -3) -1( x -3)

$\implies$(2x-1)(x-3)

$\boxed{\underline{\bold{\mathsf{Equate\: the\:factors\:to\:zero}}}}$

$\rule{200}{1}$

$\implies$(2x-1)= 0

$\implies{x = \frac{1}{2}}$

$\rule{200}{1}$

$\implies$(x-3)=0

$\implies$x = 3

$\rule{200}{1}$

as we were given p & q are its roots

as we were given p & q are its rootshere we get values of p and q

$\rule{200}{1}$

Apply squaring

$\implies{\bold{\mathsf{{(p +q)}^{2} = p^{2} + {q}^{2} + 2pq }}}$

$\implies{{(3 + \frac{1}{2}}^{2} - 2(3)(\frac{1}{2} = p^{2} + {q}^{2} }$

$\implies{{(\frac{7}{2})}^{2} - 3 =p^{2} + {q}^{2} }$

$\implies{\frac{49}{4} -3 = p^{2} + {q}^{2} }$

$\implies{ p^{2} + {q}^{2} = \frac{37}{4}}$

$\rule{200}{1}$

$\boxed{\bold{\mathsf{Formula\:Used\:are}}}$:—

• (a + b)² = a² + b² + 2ab

$\rule{200}{1}$

Answer 2

Answer:

Step-by-step explanation:

Given:

To find: p² + q²

Apply middle term splitting

x(2 x (x -3) -1( x -3)

(2x-1)(x-3)

(2x-1)= 0

(x-3)=0

x = 3

as we were given p & q are its roots

as we were given p & q are its rootshere we get values of p and q

Apply squaring

:—

• (a + b)² = a² + b² + 2ab