Question

Find the zeroes of 4root5x2 - 17x +3root5 and verify the relation between the zeroes and coefficients
of the polynomial.​

Answer 1

$\large{\underline{\bf{\green{Given:-}}}}$

✰ p(x) = 4√5x² - 17x +3√5

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the zeroes of the given polynomial.and also find the relationship between the coefficients and zeroes.

$\huge{\underline{\bf{\red{Solution:-}}}}$

$\red{\underbrace{By\: using\: quadratic\: formula}}$

$\pink{ \bf \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

✰ a = 4√5

✰ b = -17

✰ c = 3√5

Now,

,putting these values in the quadratic formula:-

$\sf \frac{ - ( - 17) \pm \sqrt{ { - 17}^{2} - 4 \times 4 \sqrt{5} \times 3 \sqrt{5} } }{2 \times 4 \sqrt{5} } \\ \\ : \implies \sf \frac{17 \pm \sqrt{289 + ( - 16 \sqrt{5}) \times 3 \sqrt{5} } }{8 \sqrt{5} } \\ \\ : \implies \sf \frac{17 \pm \sqrt{289 - 240} }{8 \sqrt{5} } \\ \\: \implies \sf \frac{17 \pm \sqrt{49} }{8 \sqrt{5} } \\ \\ : \implies \sf \frac{17 \pm7}{8 \sqrt{5} } \\ \\ : \implies \sf\boxed{ \purple{ \bf{x = \frac{24}{8 \sqrt{5} } = \frac{3}{ \sqrt{5} } }}} \\ \\ : \implies \sf \boxed{ \purple{ \bf{x = \frac{10}{8 \sqrt{5} } = \frac{5}{4 \sqrt{5} } }}}$

Now,

Relationship between the zeroes and coefficients:-

Let α = 3/√5

β = 5/4√5

α + β = -b/a

αβ = c/a

$\underbrace{\bf{\:\:sum\:of\: zeroes\:\:}}$

$: \implies \sf\frac{3}{\sqrt{5}}+\frac{5}{4\sqrt{5}}=\frac{17}{4\sqrt{5}}$

$: \implies \sf\frac{12+5}{4\sqrt{5}}=\frac{17}{\sqrt{5}}$

$: \implies \sf\frac{17}{4\sqrt{5}}=\frac{17}{4\sqrt{5}}$

Now,

$\underbrace{\bf{\:\:Product\:of\: zeroes\:\:}}$

$: \implies \sf\frac{3}{\sqrt{5}}\times\frac{5}{4\sqrt{5}}=\frac{3\sqrt{5}}{4\sqrt{5}}$

$: \implies \sf\frac{{\cancel{15}}}{{\cancel{20}}}=\frac{3}{4}$

$: \implies \sf\frac{3}{4}=\frac{3}{4}$

LHS = RHS .

Hence, Relationship is varified.

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

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

$\bold{Roots \ are \frac{3}{\sqrt5} \ and \frac{\sqrt5}{4}}$

$\bold\orange{Given:}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>4\sqrt5x^{2}-17x+3\sqrt5}$

$\bold\pink{To \ find:}$

$\bold{=>Roots \ of \ the \ polynomial.}$

$\bold\blue{Explanation}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>4\sqrt5x^{2}-17x+3\sqrt5}$

$\bold{Here, \ a=4\sqrt5, \ b=-17 \ and \ c=3\sqrt5}$

$\bold{a×c=4\sqrt5×3\sqrt5}$

$\bold{\therefore{a×c=60}}$

$\bold{Middle \ term \ should \ be \ split \ such}$

$\bold{that, \ their \ multiplication \ should \ be \ 60}$

$\bold{and \ sum \ should \ be \ -17.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{=>4\sqrt5x^{2}-17x+3\sqrt5}$

$\bold{=>4\sqrt5x^{2}-12x-5x+3\sqrt5}$

$\bold{=>4x(\sqrt5x-3)-\sqrt5(\sqrt5x-3)}$

$\bold{=>(\sqrt5x-3)(4x-\sqrt5)}$

$\bold{\therefore{x=\frac{3}{\sqrt5} \ or \frac{\sqrt5}{4}}}$

$\bold\purple{\therefore{Roots \ are \frac{3}{\sqrt5} \ and \frac{\sqrt5}{4}}}$

$\bold\green{\underline{\underline{Verification}}}$

$\bold{Let \ roots \ be \ \alpha \ and \ \beta \ be \ roots}$

$\bold{Sum \ of \ roots=\frac{-b}{a}}$

$\bold{Product \ of \ roots=\frac{c}{a}}$

__________________________________

$\bold{\alpha+\beta=\frac{3}{\sqrt5}+\frac{\sqrt5}{4}}$

$\bold{\alpha+\beta=\frac{12+5}{4\sqrt5}}$

$\bold{\therefore{\alpha+\beta=\frac{17}{4\sqrt5}...(1)}}$

$\bold{\frac{-b}{a}=\frac{-(-17)}{4\sqrt5}}$

$\bold{\therefore{\frac{-b}{a}=\frac{17}{4\sqrt5}...(2)}}$

$\bold{From \ (1) \ and \ (2)}$

$\bold{\alpha+\beta=\frac{-b}{a}}$

$\bold{\therefore{Sum \ of \ roots=\frac{-b}{a}}}$

___________________________________

$\bold{\alpha×\beta=\frac{3}{\sqrt5}× \frac{\sqrt5}{4}}$

$\bold{\therefore{\alpha×\beta=\frac{3}{4}...(3)}}$

$\bold{\frac{c}{a}=\frac{3\sqrt5}{4\sqrt5}}$

$\bold{\therefore{\frac{c}{a}=\frac{3}{4}...(4)}}$

$\bold{From \ (3) \ and \ (4)}$

$\bold{\alpha×\beta=\frac{c}{a}}$

$\bold{\therefore{Product \ of \ roots=\frac{c}{a}}}$

$\bold{}$