Question

If p and q are the zeros of the polynomial f(x) =2x²-7x+3 ,find the value of p²+q²

Answer 1

2x^2 - 7x + 3 = 2x^2 - 6x - x +3
= 2x ( x-3 ) -1 ( x-3 )
= ( 2x-1 ) ( x-3)
x= 1/2 , x= 3
p = 1/2 , q= 3


= p^2 + q^2
= (1/2)^2 + (3)^2
= 1/4 + 9
= 1+36/4
= 37/4.

Answer 2

Given: p and q are the zeros of the polynomial f(x)=2x²-7x+3.

To find: Value of p²+q²

Solution: In a quadratic polynomial

$a {x}^{2} + bx + c$

The sum of zeros = -b/a

The product of zeros = c/a

Here, in the polynomial 2x²-7x+3 :

a = 2, b = -7 and c = 3

Therefore,

Sum of zeros

= p+q

= -b/a

= -(-7)/2

= 7/2

Product of zeros

= pq

= c/a

= 3/2

Now, we know that:

${(p + q) }^{2} = {p}^{2} + {q}^{2} + 2pq$

Using the values found above:

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

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

$= > {p}^{2} + {q}^{2} = \frac{49 - 12}{4}$

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

Therefore, the value of p²+q² is 37/4.