Question

If α,β are the zeros of p(x)= ax²+bx+c,then α+β=….. (a/b , -b/a , c/a , -c/a)
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Answer 1

Answer:

α+β = -b/a

Step-by-step explanation:

α,β are the zeros of p(x) = ax² + bx + c

• x² coefficient = a
• x coefficient = b
• constant term = c

From the relation between zeroes and coefficients, we get

Sum of zeroes = -(x coefficient)/x² coefficient

α + β = -b/a

Product of zeroes = constant term/x² coefficient

αβ = c/a

Proof :

For the quadratic polynomial p(x) = ax² + bx + c , let's find the zeroes and add them.

 By quadratic formula, we know

      $\sf x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore, the two zeroes of the given polynomial are

  $\sf \dfrac{-b +\sqrt{b^2-4ac}}{2a} \ and \ \dfrac{-b -\sqrt{b^2-4ac}}{2a}$

Now, add them up to get sum of zeroes.

$\sf = \dfrac{-b +\sqrt{b^2-4ac}}{2a} + \dfrac{-b-\sqrt{b^2-4ac}}{2a} \\\\ \sf =\dfrac{-b}{2a} +\dfrac{\sqrt{b^2-4ac}}{2a}+\dfrac{-b}{2a} -\dfrac{\sqrt{b^2-4ac}}{2a} \\\\ \sf =\dfrac{-b}{2a}+\dfrac{-b}{2a} \\\\ \sf =\dfrac{-b}{2a}-\dfrac{b}{2a} \\\\ \sf = \dfrac{-b-b}{2a} \\\\ \sf = \dfrac{-2b}{2a} \\\\ \sf = \dfrac{-b}{a}$

So, sum of zeroes = -b/a