form a quadratic polynomial whose zeros are 2alpha,2ß if alpha and ß are the zeros of p(x)=x^2+12x+35
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The zeroes of the equation p(x) = x^2 + 12x + 35 are α and β as given in the question.
On solving p(x),
= x^2 + 12x + 35
= x^2 + 7x + 5x + 35 = x(x+7) + 5(x+7)
= (x+5) (x+7)
Therefore, the roots of the p(x) are -5 and -7.
Thus, from here α and β are -5 and -7 ....... (1)
And to form the quadratic equation from roots we know,
x^2 + (sum of roots) + (product of roots)
It is given that the roots of the equation we have to form are 2α and 2β, from here
= x^2 + [2α + 2β]x + 2α * 2β
= x^2 + [2(α + β)]x + 4αβ
= x^2 + 2[-5 + (-7)]x + 4(-5 * -7) ........ from (1)
= x^2 + 2(-12)x + 4(35)
= x^2 - 24x + 140
So the equation for the roots given 2α & 2β is x^2 - 24x + 140
On solving p(x),
= x^2 + 12x + 35
= x^2 + 7x + 5x + 35 = x(x+7) + 5(x+7)
= (x+5) (x+7)
Therefore, the roots of the p(x) are -5 and -7.
Thus, from here α and β are -5 and -7 ....... (1)
And to form the quadratic equation from roots we know,
x^2 + (sum of roots) + (product of roots)
It is given that the roots of the equation we have to form are 2α and 2β, from here
= x^2 + [2α + 2β]x + 2α * 2β
= x^2 + [2(α + β)]x + 4αβ
= x^2 + 2[-5 + (-7)]x + 4(-5 * -7) ........ from (1)
= x^2 + 2(-12)x + 4(35)
= x^2 - 24x + 140
So the equation for the roots given 2α & 2β is x^2 - 24x + 140
Vinayak2001Rana:
zeros are 2(alpha) and 2(ß)
not alpha and ß
There is a equation given already with its zeroes, those are alpha and beta. The question asks you to form the equation of 2alpha and 2beta, after you find the roots and then you can put them to find the answer on it
ok i got it thanx
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