Question

if a and b are the zeroes of p (x) = ax^2-2x+3a and 1/a + 1/b =1 ,then find a ?​

Answer 1

✪AnSwEr

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

• $\tt\alpha{\:and}\beta$ are zeroes of polynomial
• p(x)a=ax²-2x+3a
• $\tt\dfrac{1}{\alpha}{+}\dfrac{1}{\beta}=1$

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

• Value of a

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

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

Then,

$\tt\alpha{+}\beta\frac{-b}{a}=\frac{2}{a}$

&

$\tt\alpha{\times}\beta{=}\frac{c}{a}=\frac{3a}{a}=3$

Solving

$\implies\tt\dfrac{1}{\alpha}{+}\dfrac{1}{\beta}=1$

$\implies\tt\dfrac{\alpha+\beta}{\alpha\times\beta}=1$

¶utting values

=>{2/a}/3=1

=>2/a=3

=>2=3a

=>a=2/3

polynomial be

p(x)=2x²-6x+6

Additional Information

Here,

a=2

b=-6

C=6

☞

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{6}{2}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{6}{2}$

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

Hence,relation verified