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Standard VI

Mathematics

Formation of Algebraic Expressions

If P = 2y2 ...

Question

If $P = 2 y^{2} - 5 - 2 y, Q = y^{2} - 3 y + 6$ and $R = y^{2} - 4 y + 2$ , find $P + 2 Q + 2 R$ .

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Solution

Given,
$P = 2 y^{2} - 5 - 2 y, Q = y^{2} - 3 y + 6$ and $R = y^{2} - 4 y + 2$ .
Now,
$P + 2 Q + 2 R$ $= P + 2 (Q + R)$
$= 2 y^{2} - 5 - 2 y + 2 (y^{2} - 3 y + 6 + y^{2} - 4 y + 2)$
$= 2 y^{2} - 5 - 2 y + 2 (2 y^{2} - 7 y + 8)$
$= 2 y^{2} - 2 y - 5 + 4 y^{2} - 14 y + 16$
$= 6 y^{2} - 16 y + 11$ .

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Similar questions

\frac{y}{y - 2} + \frac{1}{y + 2} + \frac{3 y}{4 - y^{2}}

= .

Q. The line

x + 2 y + 3 = 0

cuts the circle

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

at points

P

and

Q

and the line

2 x + 3 y + λ = 0

cuts the circle

x^{2} + y^{2} + 6 x + 2 y - 7 = 0

at points

R

and

S .

P,

Q,

R

and

S

are concyclic, then the value of

λ

Q. If

2 Q + 3 P = \frac{3 P - Q}{2 R}

, express P in terms of Q and R. Find the value of P, when

R = - 2

and

Q = 1

Match the column

$\begin{matrix} Equation & Name of the curve 1) x^{2} - 2 x - y - 3 = 0 & P) Circle 2) x^{2} + 3 x y + 2 y^{2} - x - 4 y - 6 = 0 & Q) Parabola 3) x^{2} + y^{2} - 20 = 0 & R) Ellipse 4) 7 x^{2} + 7 y^{2} + 2 x y + 10 x - 10 y + 7 = 0 & S) Hyperbola 5) 6 x^{2} - x y - y^{2} - 23 x + 4 y + 15 = 0 & T) Pair of straight lines \end{matrix}$

Find the sum of the following polynomials

y^{2} + 2 y - 5; y^{3} + 2 y^{2} + 3 y + 4; y^{3} + 7 y - 2

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If P=2y2−5−2y,Q=y2−3y+6 and R=y2−4y+2, find P+2Q+2R.

Given, P=2y2−5−2y,Q=y2−3y+6 and R=y2−4y+2.Now,P+2Q+2R =P+2(Q+R)=2y2−5−2y+2(y2−3y+6+y2−4y+2)=2y2−5−2y+2(2y2−7y+8)=2y2−2y−5+4y2−14y+16=6y2−16y+11.

If $P = 2 y^{2} - 5 - 2 y, Q = y^{2} - 3 y + 6$ and $R = y^{2} - 4 y + 2$ , find $P + 2 Q + 2 R$ .

Given,
$P = 2 y^{2} - 5 - 2 y, Q = y^{2} - 3 y + 6$ and $R = y^{2} - 4 y + 2$ .
Now,
$P + 2 Q + 2 R$ $= P + 2 (Q + R)$
$= 2 y^{2} - 5 - 2 y + 2 (y^{2} - 3 y + 6 + y^{2} - 4 y + 2)$
$= 2 y^{2} - 5 - 2 y + 2 (2 y^{2} - 7 y + 8)$
$= 2 y^{2} - 2 y - 5 + 4 y^{2} - 14 y + 16$
$= 6 y^{2} - 16 y + 11$ .