If $P = {3 x}^{2} + 7 x + 8, Q = {2 x}^{2} + x - 9$ and $R = - {5 x}^{2} - 8 x + 1$ , show that $P + Q + R = 0$

Open in App

Solution

$P + Q + R = 3 x^{2} + 7 x + 8 + 2 x^{2} + x - 9 - 5 x^{2} - 8 x + 1$

$= (3 x^{2} + 2 x^{2} - 5 x^{2}) + (7 x + x - 8 x) + (8 - 9 + 1)$

$= 0 + 0 + 0 = 0$

Hence proved.

Suggest Corrections

Similar questions

P = 7 x^{2} + 5 x y - 9 y^{2}, Q = 4 y^{2} - 3 x^{2}

R = 4 x^{2} + x y + 5 y^{2}

P + Q + R = 8 x^{2} + 6 x y

a

(x^{2} + x + 2)^{2} - (a - 3) (x^{2} + x + 2) (x^{2} + x + 1) + (a - 4) (x^{2} + x + 1)^{2} = 0

(p, \frac{q}{r}]

p + q + r

a

(x^{2} + x + 2)^{2} - (a - 3) (x^{2} + x + 2) (x^{2} + x + 1) + (a - 4) (x^{2} + x + 1)^{2} = 0

(p, \frac{q}{r}]

p + q + r

5^{- p} = 4^{- q} = 20^{r}

\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0

Join BYJU'S Learning Program