Math, asked by Kausikivarma528, 9 months ago

Find the zeros
v²+4√3v-15 verification

Answers

Answered by Abhishek474241
7

AnSwEr

{\tt{\red{\underline{\large{Given}}}}}

  • A polynomial
  • V²+4√3v-15

{\sf{\green{\underline{\large{To\:Find}}}}}

  • Factors of the polynomial
  • Relationship between cofficient

{\sf{\pink{\underline{\Large{Explanation}}}}}

V²+4√3v-15

we have to spilt the middle term in such a way that the product become -15and sum become 4√3

V²+4√3v-15

=>V²+5√3V-√3V-15

=>V(v+5√3)-√3(v+5√3)

=>(v-√3) (v+5√3)

Additional Information

Let the zeroes of the polynomial be\tt\alpha{and}\beta

Then,

\rightarrow\tt\alpha{+}\beta{=}\frac{-b}{a}

&

\rightarrow\tt\alpha{\times}\beta{=}\frac{c}{a}

Here,

a=1

b=4√3

C=-15

\rightarrow\tt\alpha{+}\beta{=}\dfrac{4\sqrt{3}}{1}

\rightarrow\tt\alpha{+}\beta{=}\dfrac{-(Cofficient\:of\:V)}{Cofficient\:of\:V^2}=

&

\rightarrow\tt\alpha{\times}\beta{=}\dfrac{-15}{1}

\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}

Hence,relation verified

Answered by Anonymous
1

Hello,

ANSWER : -

\longmapsto\tt\bold{{v}^{2}+4\sqrt{3}v-15}

→Now, Split the middle term in the way that sum becomes 4√3 and Product -15 ..

\longmapsto\tt\bold{{v}^{2}+4\sqrt{3}v-15}\\ \\ \longmapsto\tt\bold{{v}^{2}+5\sqrt{3}v-\sqrt{3}v-15}\\ \\ \longmapsto\tt\bold{v(v+5\sqrt{3})-\sqrt{3}\:\:(v+5\sqrt{3})}\\ \\ \longmapsto\tt\bold{(v-\sqrt{3}\:\: (v+5\sqrt{3})}

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