Question

if A,Bate the roots of 2x²-3x+5=0then find A²+B²​

Answer 1

Step-by-step explanation:

sum of roots = A+B =

$\frac{ - b}{a}$

= 3/2

AB = Product of roots = c/a = 5/2

${a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab \\ = {( \frac{3}{2}) }^{2} - 2 \times \frac{5}{2} \\ = \frac{9}{4} - 5 \\ = \frac{ - 11}{4}$

Answer 2

Question :-

• If A, B are the roots of 2x²- 3x + 5 = 0, then find A²+B².

Answer

Given :-

• A and B are the rots of 2x²- 3x + 5 = 0

To Find :-

• The value of A²+B².

Concept used :-

• If α,β be two zeroes of the quadratic equation ax² + bx + c = 0, then

$\bf \:Sum \: of \: zeroes = \alpha + \beta = - \dfrac{b}{a}$

$\bf \:Product \: of \: zeroes = \alpha \beta = \dfrac{c}{a}$

$\bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta$

Solution :-

Since A and B are zeroes of 2x²- 3x + 5 = 0, then

$\bf\implies \:A + B = - \dfrac{( - 3)}{2} = \dfrac{3}{2}$

$\bf\implies \:AB = \dfrac{5}{2}$

Consider, A² + B²

$\bf\implies \:A²+B² = {(A + B)}^{2} - 2AB$

$\bf\implies \:A²+B² = {( \dfrac{3}{2} )}^{2} - 2 \times \dfrac{5}{2}$

$\bf\implies \:A²+B² = \dfrac{9}{4} - 5$

$\bf\implies \:A²+B² = - \dfrac{11}{4}$

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