Question

If p, q are zeroes of polynomial 2x2 -7x +3 find the value of p2 + q2

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{p^{2}+q^{2}=\frac{37}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {2x}^{2} - 7x + 3 = 0 \\ \\ \tt: \implies p \: and \: q \: are \: zeroes \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies {p}^{2} + {q}^{2} =?$

• According to given question :

$\tt \circ \: {2x}^{2} - 7x + 3 = 0 \\ \\ \tt \circ \: a = 2 \: \: \: \: \: \: b = - 7 \: \: \: \: \: \: c = 3 \\ \\ \bold{For \: sum \: of \: zeroes :} \\ \tt: \implies p + q = \frac{ - b}{a} \\ \\ \tt: \implies p + q = \frac{ - ( - 7)}{2} \\ \\ \green{\tt: \implies p + q = \frac{7}{2} } \\ \\ \bold{For \: product \: of \: zeroes : } \\ \tt: \implies p \times q = \frac{c}{a} \\ \\ \green{\tt: \implies pq = \frac{3}{2}} \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies {p}^{2} + {q}^{2} = {(p + q)}^{2} - 2pq \\ \\ \tt: \implies { p}^{2} + {q}^{2} = (\frac{7}{2} )^{2} - 2 \times \frac{3}{2} \\ \\ \tt: \implies { p}^{2} + {q}^{2} = \frac{49}{4} - 3 \\ \\ \tt: \implies { p}^{2} + {q}^{2} = \frac{49 - 12}{4} \\ \\ \green{\tt: \implies { p}^{2} + {q}^{2} = \frac{37}{4} }$

Answer 2

Aɴꜱᴡᴇʀ

☞ p² + q² = 9.25

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Gɪᴠᴇɴ

➳ P and Q are the zeros of the polynomial 2x²- 7x + 3

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Tᴏ ꜰɪɴᴅ

➤ p² + q² ?

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Sᴛᴇᴘꜱ

❍ First let's find the value of p and q

$\leadsto \sf 2 {x}^{2} - 7x + 3 = 0 \\ \\ \leadsto \sf2 {x}^{2} - 6x - 1x + 3 = 0 \\ \\ \leadsto \sf 2x(x - 3) - 1(x - 3) = 0 \\ \\ \leadsto{} \sf(2x - 1)(x - 3) = 0$

Equating each seperately with 0,

$\longrightarrow \sf(2x - 1) = 0 \\ \\ \longrightarrow{} \sf \red{x = \frac{1}{2}}$

$\longrightarrow \sf(x -3 ) = 0 \\ \\ \longrightarrow \red{ \sf x = 3}$

Therefore,

$\boxed{ \bigstar{} \sf{} \: \: p = \frac{1}{2} } \\ \\ \boxed{\bigstar{} \: \: \sf{}q = 3}$

So substituting in p² +q²

$\dashrightarrow{} \sf {p}^{2} + {q}^{2} = ( \frac{1}{2 }{)}^{2} + {(3)}^{2} \\ \\ \dashrightarrow{} \sf \frac{1}{4} + 9 \\ \\ \dashrightarrow \sf \frac{1 + 36}{4} \\ \\ \sf\pink{\dashrightarrow \frac{37}{4} \: \: or\:\: 9.25 }$

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