Question

If p and q are the zeros of the polynomial f(x) =2x²-7x+3 ,find the value of p²+q²

Answer 1

☯GIVEN :

• $\bf {2x^2-7x + 3}$ , where p and q are the zeros of the given polynomial.

☯TO FIND :

• The value of p²+q².

➲SOLUTION :

★ Finding p and q :

$:\implies \sf{2x^2 -7x + 3 = 0}$

• By middle term splitting :

$:\implies \sf{2x^2-6x-x+3= 0}$

$:\implies \sf{2x \bigg(x -3 \bigg)-1\bigg(x-3\bigg) = 0}$

$:\implies \sf{\bigg(x-3 \bigg) \bigg(2x -1\bigg) = 0}$

$:\implies \sf{\bigg(x-3 \bigg) = 0 \: , \: \bigg(2x -1 \bigg) = 0}$

$:\implies\sf{x = 0 + 3 \: , \: 2x = 0+1}$

$:\implies\sf{x = 3 \: , \: 2x = 1}$

$\leadsto\underline{\huge{\boxed{\blue{\bf{x = 3\: , \: x = \dfrac {1}{2}}}}}}$

Here,

• p = 3 and q = $\bf {\dfrac {1}{2}}$

★ Finding p²+q² :

$: \leadsto \sf{ p^2 + q^2}$

$: \leadsto \sf{ (3)^2 + {\left(\dfrac{1}{2} \right)}^{2}}$

$: \leadsto \sf{ 9 +\dfrac{1}{4} }$

$: \leadsto \sf{ \dfrac{36 + 1}{4} }$

$\mapsto \underline{ \huge{ \purple{\boxed{\bf{ \dfrac{37}{4} }}}}}$

$\huge{\pink{\therefore}}$ The value of p²+q² is $\bf{ \dfrac{37}{4}}$.