Question

plz give answer with solution !​

Answer 1

Step-by-step explanation:

$= > (x - 2)(x - \frac{1}{2} ) \: \: \: \: \: \: \: \: \: \: \\ = > {x }^{2} - 2x - \frac{1}{2}x + 1 \: \: \: \: \: \\ = > 2 {x}^{2} - 4x - x + 2 \: \: \: \: \: \: \\ = > 2x {}^{2} - 5x + 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = > ( - 2x {}^{2} ) + 5x + ( - 2)$

According to the question,

$= > p {x}^{2} + 5x + r$

By comparing....

P = - 2

R = - 2

P = R

Hence Proved

Answer 2

Answer:

Question :-

$\mapsto \sf If\: both\: (x - 2)\: and\: \bigg(x - \dfrac{1}{2}\bigg)\: are\: factors\: of\: px^2 + 5x + r, prove\: that\: p =\: r\: .$

Solution :-

Given :

$\leadsto \sf (x - 2)\: and\: \bigg(x - \dfrac{1}{2}\bigg)\: are\: factors\: of\: px^2 + 5x + r\: .$

when, (x - 2) = 0 :

$\implies \sf x - 2 =\: 0$

$\implies \sf\bold{\green{x =\: 2}}$

So, by putting x = 2 we get,

$\implies \bf{px^2 + 5x + r =\: 0}$

$\implies \sf p(2)^2 + 5(2) + r =\: 0$

$\implies \sf p(4) + 10 + r =\: 0$

$\implies \sf 4p + 10 + r =\: 0$

$\implies \sf\bold{\purple{4p + r =\: - 10\: ------\: (Equation\: No\: 1)}}$

Again, when, (x - 1/2) = 0 :

$\implies \sf x - \dfrac{1}{2} =\: 0$

$\implies \sf\bold{\green{x =\: \dfrac{1}{2}}}$

So, by putting x = ½ we get,

$\implies \bf{px^2 + 5x + r =\: 0}$

$\implies \sf p\bigg(\dfrac{1}{2}\bigg)^2 + 5\bigg(\dfrac{1}{2}\bigg) + r =\: 0$

$\implies \sf p\bigg(\dfrac{1}{4}\bigg) + \dfrac{5}{2} + r$

$\implies \sf \dfrac{p}{4} + \dfrac{5}{2} + r =\: 0$

$\implies \sf \dfrac{p + 10 + 4r}{4} =\: 0$

By doing cross multiplication we get,

$\implies \sf p + 10 + 4r =\: 0(4)$

$\implies \sf p + 10 + 4r =\: 0$

$\implies \sf p + 4r =\: - 10$

$\implies \sf\bold{\purple{p =\: - 10 - 4r\: ------\: (Equation\: No\: 2)}}$

Now, by putting the value of p in the equation no 1 we get,

$\implies \sf 4p + r =\: - 10$

$\implies \sf 4(- 10 - 4r) + r =\: - 10$

$\implies \sf - 40 - 16r + r =\: - 10$

$\implies \sf - 40 - 15r =\: - 10$

$\implies \sf - 15r =\: - 10 + 40$

$\implies \sf - 15r =\: 30$

$\implies \sf r =\: \dfrac{\cancel{30}}{- \cancel{15}}$

$\implies \sf r =\: \dfrac{2}{- 1}$

$\implies \sf\bold{\red{r =\: - 2}}$

Again, by putting the value of r in the equation no 1 we get,

$\longrightarrow \sf 4p + r =\: - 10$

$\longrightarrow \sf 4p + (- 2) =\: - 10$

$\longrightarrow \sf 4p - 2 =\: - 10$

$\longrightarrow \sf 4p =\: - 10 + 2$

$\longrightarrow \sf 4p =\: - 8$

$\longrightarrow \sf p =\: \dfrac{- \cancel{8}}{\cancel{4}}$

$\longrightarrow \sf p =\: \dfrac{- \cancel{4}}{\cancel{2}}$

$\longrightarrow \sf p =\: \dfrac{- 2}{1}$

$\longrightarrow \sf\bold{\red{p =\: - 2}}$

Hence, we have :

• p = - 2
• r = - 2

$\leadsto \sf p =\: r$

$\leadsto \sf\bold{- 2 =\: - 2}$

$\leadsto \bf{L.H.S =\: R.H.S}$

$\mapsto \sf\boxed{\bold{\pink{Hence\: Proved}}}$