Let $p$ and $q$ be the degree and the number of zeros of the polynomial $g (x) = x^{2} - 6 x + 9$ then find $p + q - p q$

Open in App

Solution

Degree of the polynomial $=$ highest power of the polynomial $= 2$
$g (x) = x^{2} - 6 x + 9$
$= x^{2} - 2 \times 3 x + 3^{2}$
$= {(x - 3)}^{2}$
$∴$ number of zeros $= 2$
$∴ q = 2$
Hence $p + q - p q = 2 + 2 - 2 \times 2 = 4 - 4 = 0$

Suggest Corrections

Similar questions

Find the zero of the polynomial x²-bxy + cy²where -b =p-q and c =-pq

Q. If

α

and

β

are the zeros of the polynomial

(x - p) (x - q) - r

, then

p

and

q

would be the zeros of the polynomial

For hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ , and circle $x^{2} + y^{2} = 49$ , Let $P, Q, R, S$ be concyclic points P in 1st and Q in 2nd qudrant.

Find slope of $P Q$ .

‘-p’ and ‘q’ are the zeroes of the polynomial
$x^{2} - b x + c$ .
The zeroes of the polynomial $x^{2} - b x y + c y^{2}$ are

Q. Find the zero of the polynomial in each of the following cases.

f (x) = p x + q, p \neq 0, p q

are real numbers

Join BYJU'S Learning Program

Let p and q be the degree and the number of zeros of the polynomial g(x)=x2−6x+9 then find p+q−pq

Degree of the polynomial=highest power of the polynomial=2g(x)=x2−6x+9=x2−2×3x+32=(x−3)2∴ number of zeros=2∴q=2Hence p+q−pq=2+2−2×2=4−4=0

Let $p$ and $q$ be the degree and the number of zeros of the polynomial $g (x) = x^{2} - 6 x + 9$ then find $p + q - p q$

Degree of the polynomial $=$ highest power of the polynomial $= 2$
$g (x) = x^{2} - 6 x + 9$
$= x^{2} - 2 \times 3 x + 3^{2}$
$= {(x - 3)}^{2}$
$∴$ number of zeros $= 2$
$∴ q = 2$
Hence $p + q - p q = 2 + 2 - 2 \times 2 = 4 - 4 = 0$