Question

Q1.If r+s=6 and rs = -27, then find the value of r²+s²​.

Answer 1

Answer:

90

Explanation:

r+s =9

rs = -27

-3+9 = 9

-3*9 = 27

Answer 2

Explanation:

Given

r+s= 6 and rs=-27

To Find

We have to find the value of r²+s²

$\sf\huge\bold{\underline{\underline{{Solution}}}} </p><p>Solution</p><p>$

➙r+s=6

➙rs= -27

Here,an identity will use

look at into given values

r+s represent a+b & rs represent ab

and have to find r²+s²

So, we can use this identity

➙(a+b)²=a²+b²+2ab

Now, make identity according to the question:

➙(r+s)²=r²+s²+2rs

Put given values into identity

➙(6)²= r²+s²+2(-27)

➙36=r²+s²-54

➙36+54=r²+s²

➙r²+s²=90

Hence,the value of r²+s² is 90.

More identities:

$\sf\bold{\boxed{ {(x -y)}^{2} = {x}^{2} + {y}^{2} - 2xy}} </p><p>(x−y) </p><p>2</p><p> =x </p><p> ^{2} </p><p> +y </p><p> ^{2} </p><p> −2xy$

$\sf\bold{\boxed{ {x}^{3} + {y}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}} </p><p>x </p><p> ^{2} </p><p> +y </p><p> ^{2} </p><p> =x </p><p> ^{2} </p><p> +y </p><p> ^{2} </p><p> +3xy(x+y)$

$\sf\bold{\boxed{(x + y)(x + z) = {x}^{2} + (y+ z)x + yz}} </p><p>(x+y)(x+z)=x </p><p> ^{2} </p><p> +(y+z)x+yz</p><p>$

$\sf\bold{\boxed{ {(x +y)}^{3} = {x}^{3} + {y}^{3} +3xy[x+y]}} </p><p>(x+y) </p><p> ^{2} </p><p> =x </p><p> ^{2} </p><p> +y </p><p> ^{2} </p><p> +3xy[x+y]</p><p> </p><p>$

$\sf\bold{\boxed{(x + a)(x - b) = {x}^{2} + (a - b)x - ab}} </p><p>(x+a)(x−b)=x </p><p> ^{2} </p><p> +(a−b)x−ab</p><p>$

$\sf\bold{\boxed{(x - a)(x - b) = {x}^{2} - (a + b)x + ab}} </p><p>(x−a)(x−b)=x </p><p> ^{2} </p><p> −(a+b)x+ab$