Question

if a and b are the zeros of quadratic polynomial p(x)= x^2 - 4x + 3 find the value of a^4 b^2 + a^2 b^4​

Answer 1

${\underline{\sf{Question}}}$

If a and b are the zeros of quadratic polynomial $\sf\:p(x)=x{}^{2}-4x+3$find the value $\sf\:a{}^{4}b{}^{2}+a{}^{2}b{}^{4}$.

${\underline{\sf{Solution}}}$

We have ,$\sf\:p(x)=x{}^{2}-4x+3$

Since , a and b are the zeros of quadratic polynomial p(x)

$\sf \implies \: p(x) = 0$

$\sf \implies \: x {}^{2} - 4x + 3 = 0$

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

$\sf \implies(x - 3)(x -1 ) = 0$

$\sf \implies \: x = 3 \: or \: 1$

Therefore , a and b are 3 and 1

We have to find the value of

$\sf {a}^{4} b {}^{2} + a {}^{4} b {}^{2}$

$\sf = {a}^{2} {b}^{2} (a {}^{2} + b {}^{2} )$

put the values of a and b

$\sf = 3 {}^{2} \times 1 {}^{2} (3 {}^{2} + 1 {}^{2} )$

$= \sf9(9 + 1)$

$= \sf9 \times 10 = 90$

Answer 2

$\LARGE{\sf{\purple{Given}}}$

• $\sf{Quadratic\: polynomial\: = {x}^{2} - 4x + 3}$

• $\sf{Zeros\: of \: quadratic\: polynomial\: = \: a \: and \: b }$

$\LARGE{\sf{\purple{To\: Find}}}$

• $\sf{Value\: of \: {(a)}^{4}\times{(b)}^{2} + {(a)}^{2}\times{(b)}^{4} }$

$\LARGE{\sf{\purple{Solution }}}$

We know that if any quadratic polynomial x² + bx +c = 0 have roots p and q , then

$\underline{\sf{\implies p \times q = \frac{c}{a}}}$

$\underline{\sf{\implies p + q = - \frac{b}{a} }}$

So now according to the question we have to find a⁴.b² + a².b⁴ .

$\sf{\implies a + b = \frac{-b}{a}, \: here \: b = -4 \: and \: a = 1}$

${\underline{\sf{\implies a + b = -\frac{-4}{1} = 4 }}}$

$\sf{\implies a.b = \frac{c}{a} , \: here \: c = 3 \: and \: a = 1}$

$\underline{\sf{\implies a.b = \frac{3}{1} = 3 }}$

Now taking a⁴.b² + b⁴.a² →

$\sf{\implies {a}^{4}.{b}^{2} + {a}^{2}.{b}^{4} }$

$\sf{Taking\: {a}^{2}.{b}^{2} \: common }$

$\sf{\implies {a}^{2}.{b}^{2}[ {a}^{2} + {b}^{2}\:\:\:\: eq\:1st}$

$\sf{\implies {a}^{2}+{b}^{2} \: can \: be \: written\: as {(a+b)}^{2} - 2ab }$

$\sf{\implies {a}^{2}.{b}^{2} \: can \: be \: written\: as \: {(a.b)}^{2}}$

$\underline{\sf{\implies {( a.b)}^{2} = {3}^{2} → \: 9}}$

$\sf{\implies {(a+b)}^{2} - 2ab = {(4)}^{2} - 2(3) }$

$\underline{\sf{\implies {(a+b)}^{2} - 2ab = 16 -6 → \: 10}}$

Now putting these values in eq 1st :-

$\sf{\implies {(a.b)}^{2} [ {(a+b)}^{2}-2ab ] = 9 [10] }$

$\underline{\sf{\purple{\implies {a}^{4}.{b}^{2} + {b}^{4} .{a}^{2} = 90}}}$