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​If α,βα,β α,β​ are the roots of px2+qx+r=0px² + qx + r = 0px2+qx+r=0 then α2+β2=α^2+β^2=α2+β2=.....?
Question

​If α,βα,β ​ are the roots of px2+qx+r=0px² + qx + r = 0 then α2+β2=α^2+β^2=.....?

A.

(q2+2pr)p2\frac{(q^2+2pr)}{p²} ​​

B.

(q22pr)p2\frac{(q^2-2pr)}{p²}​​

C.

2pq2pq ​​

D.

2pq-2pq ​​

Correct option is B

Given:

α and β\betaβ are the roots of the quadratic equation: px2+qx+r=0px^2 + qx + r = 0 

α2+β2=?\alpha^2 + \beta^2 = ? 

Formula Used: 

(a+b)2=a2+b2+2ab(a +b)^2 = a^2+b^2+2ab​​

Solution:

For the quadratic equation px2+qx+r=0px^2 + qx + r = 0 the sum and product of the roots are: 

Sum of roots = α+β=qp\alpha + \beta = -\frac{q}{p} 

Product of roots = αβ=rp\alpha \beta = \frac{r}{p} 

​We use the identity:

α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta 

Substitute the expressions for α + β and αβ:

α2+β2=(qp)22(rp)\alpha^2 + \beta^2 = \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) 

α2+β2=q2p22rp\alpha^2 + \beta^2 = \frac{q^2}{p^2} - \frac{2r}{p} 

α2+β2=q22prp2\alpha^2 + \beta^2 = \frac{q^2-2pr}{p^2}  

Thus, the expression for α2+β2\alpha^2 + \beta^2 is :  (q22pr)p2 \frac{(q^2-2pr)}{p^2} .​

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