If alpha and beta are the zeros of the quadratic polynomial

  1. Zeros of Quadratic Polynomial
  2. If α and β are the zeroes of the quadratic polynomial fx=a x2+b x+c, then evaluate :i α βii 1 α 1 βiii 1 α+1 β 2 αβiv α2β αβ2v α4+β4vi 1 a α+ b +1 a β+ bvii β a α+b+α a β+bviii a α 2 β+β 2 α+b αβ+βα
  3. If α And β Are the Zeros of the Quadratic Polynomial F(X) = X2 − 1, Find a Quadratic Polynomial Whose Zeroes Are `(2alpha)/Beta" and "(2beta)/Alpha`
  4. If $\alpha$ and $\beta$ are zeros of the polynomial $4x^2


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Zeros of Quadratic Polynomial

In this mini-lesson, we will explore the world of zeros of a quadratic polynomial. We will try to understand the methods to find the zeros or roots of a quadratic polynomial, observe the roots graph, and discover other interesting aspects of it. You can check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page. Before we get started, let's have a look at a simple question. Have you ever observed the path of a ball when it is thrown in the air? Its trajectory can be modeled by a quadratic polynomial. The word " Quadratic"is derived from the word " Quad"which means square. In other words, a quadratic polynomialis a“ polynomial function of degree 2” There are many scenarios where quadratic polynomialsare used. In fact, when a rocket is launched, its path is described by the zero of aquadratic polynomial. Let's go ahead and learn more about thistopic now. Lesson Plan What Do You Mean By Zeros of a Quadratic Polynomial? Quadratic Polynomial:Definition A Zeros of Quadratic Polynomial:Definition A Consider the polynomial: \(p(x)=x^2-3x-4\) Let's find the value of the polynomial at \(x=-1\) by substituting -1 for \(x\) in \(p(x)\). \[\begin\] So, here \(x=-1\) is a zero of the quadratic polynomial\(p(x)=x^2-3x-4\). How To Find Zeros of a Quadratic Polynomial? There are two methods to find the zeros of a quadratic polynomial: • • Let us consider the quadratic polynomial \(2x^2+x-3=0\)...

If α and β are the zeroes of the quadratic polynomial fx=a x2+b x+c, then evaluate :i α βii 1 α 1 βiii 1 α+1 β 2 αβiv α2β αβ2v α4+β4vi 1 a α+ b +1 a β+ bvii β a α+b+α a β+bviii a α 2 β+β 2 α+b αβ+βα

(i) Given α and are the zeros of the quadratic polynomial f (x) = We have, Substituting and then we get, Hence, the value of is . (ii) Given α and are the zeros of the quadratic polynomial f (x) = We have, Substituting and then we get, By taking least common factor we get, Hence the value of is . Given α and are the zeros of the quadratic polynomial f (x) = We have, By cross multiplication we get, By substituting and we get , Hence the value of is . Given α and are the zeros of the quadratic polynomial f (x) = We have, By taking common factor we get, By substituting and we get , Hence the value of is . Given α and are the zeros of the quadratic polynomial f (x) = We have, By substituting and we get , By taking least common factor we get α 4 + β 4 = - b a 2 - 2 × c a 2 - 2 × c a 2 = b 2 a 2 - 2 c a 2 - 2 × c a 2 = b 2 - 2 a c a 2 2 - 2 × c 2 a 2 = b 2 - 2 a c 2 a 4 - 2 × c 2 a 2 = b 2 - 2 a c 2 - 2 c 2 a 2 a 4 Hence the value of α 4 + β 4 is b 2 - 2 a c 2 - 2 c 2 a 2 a 4 . (vi) Since α and are the zeros of the quadratic polynomial = We have, By substituting and we get , Hence, the value of is . (vii) Since α and are the zeros of the quadratic polynomial = We have, By substituting and we get , Hence, the value of is . (viii) Since α and are the zeros of the quadratic polynomial = We have, By substituting and we get , Hence, the value of is .

If α And β Are the Zeros of the Quadratic Polynomial F(X) = X2 − 1, Find a Quadratic Polynomial Whose Zeroes Are `(2alpha)/Beta" and "(2beta)/Alpha`

Since αand βare the zeros of the quadratic polynomialf(x) = x 2− 1 The roots are αand β `alpha+beta="-coefficient of x"/("coefficient of "x^2)` `alpha+beta=0/1` `alpha+beta=0` `alphabeta="constant term"/("coefficient of "x^2)` `alphabeta=(-1)/1` `alphabeta=-1` Let S and P denote respectively the sum and product of zeros of the required polynomial. Then, `S=(2alpha)/beta+(2beta)/alpha` Taking least common factor we get, `S=(2alpha^2+2beta^2)/(alphabeta)` `S=(2(alpha^2+beta^2))/(alphabeta)` `S=(2[(alpha+beta)-2alphabeta])/(alphabeta)` `S=(2[(0)-2(-1)])/-1` `S=(2[-2(-1)])/-1` `S=(2xx2)/-1` `S=4/-1` S = -4 `P=(2alpha)/betaxx(2beta)/alpha` P = 4 Hence, the required polynomial f(x) is given by, f(x) = k(x 2 - Sx + P) f(x) = k(x 2 -(-4)x + 4) f(x) = k(x 2 +4x +4) Hence, required equation isf(x) = k(x 2 +4x +4)Where kis any non zero real number.

If $\alpha$ and $\beta$ are zeros of the polynomial $4x^2

I am utterly stuck at this question which I am sure has a simple solution, but I just can't seem to be able to see it. Any help would be greatly appreciated. If $\alpha$ and $\beta$ are zeros of the polynomial $4x^2 -x -4$, find the quadratic polynomial whose zeros are $\frac$. Thank you! $\begingroup$ @ClaudeLeibovici I have tried splitting the middle term to find the zeroes of the first quadratic, but thats the stage where I get stuck. Product being 16, and the sum being -1? I can't find the suitable numbers. Otherwise, plan was to find the zeros and input the values into the given fractions. $\endgroup$ Let $y=\frac.$$ Now all you have to do is express $4y^2-y-4$ in terms of $x$. You will get a rational fraction, but it can only cancel when its numerator does; hence this numerator is your polynomial. You have $4x^2-x-4 = (x-\alpha)(x-\beta)$. If you multiply the last bit out, you get expressions for $\alpha+\beta$ and $\alpha\beta$. The quadratic you're looking for is $(x-1/2\alpha)(x-1/2\beta)$. If you multiply this out and clear fractions, the expressions you found above should give you your answer.

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