Question

if alpha and beta are the zeroes of the quadratic polynomial f(x)=ax²+bx+c then evaluate alpha square + beta square
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Answer 1

Answer:

Answer:

Step-by-step explanation:

Question:

If alpha and beta are the zeroes of the quadratic polynomial f(x)=ax²+bx+c then evaluate alpha square + beta square

Given:

• Alpha and Beta are the zeroes of the quadratic polynomial f(x)=ax²+bx+c

To find:

Alpha square + Beta square

Let:

• $\alpha$ = Alpha
• $\beta$ = Beta

​Solution:

We know that,

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

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

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

$\implies \dfrac{(-b)^2}{(a)^2} - \dfrac{2c}{a}\\\implies \dfrac{b^2-2ac}{a^2}$

Answer 2

Given:-

• f(x) = ax²+bx+c = 0

To find:-

• α² + ß²

α+β = $\sf\dfrac{-b}{a}$

αß = $\sf\dfrac{c}{a}$

Identity:-

$\underline{\boxed{\sf α²+ß² = (α+ß)²-2αß}}$

Solution:-

$\implies$ (α+ß)²-2αß

$\implies$ ${\bigg(\sf\dfrac{-b}{a}\bigg)}^{2}$ - $2 \times \bigg(\sf\dfrac{c}{a}\bigg)$

$\implies$ $\sf\dfrac{b²}{a²}$ - $\sf\dfrac{2c}{a}$

$\implies$ $\sf\dfrac{b²-2ac}{a²}$