## msingh Group Title SOLVE if alpha and beta are zeroes of x^2+7x+K. find value of k one year ago one year ago

1. hartnn Group Title

anything else given for alpha and beta ? or you need k in terms of alpha and beta ?

2. myko Group Title

Hint: $$(x-\alpha)(x-\beta)=x^2+7x+K$$

3. msingh Group Title

question has not sufficient data

4. myko Group Title

it has

5. hartnn Group Title

well, product of roots(zeros) for ax^2+bx+c =0 is just c/a so here, k/1 = alpha*beta , gives k=alpha*beta

6. mathslover Group Title

$$\alpha + \beta = -7$$ and $$\alpha \times \beta = k$$

7. myko Group Title

$$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=x^2+7x+K$$\) can you finish this

8. hartnn Group Title

still, you'll not get a value for k, you'll be able to express k in terms of either alpha or beta.

9. myko Group Title

yes.

10. myko Group Title

good enough

11. msingh Group Title

@hartnn is absolutely correct

12. hartnn Group Title

so is @myko

13. mathslover Group Title

I am trying to say something but checking my method first. It seems somewhat different .

14. mathslover Group Title

$$(\alpha + \beta)^2 = 49$$ $$\alpha^2 + \beta^2 + 2(k) = 49$$ $$\alpha^2 + \beta^2 = 49-2k$$ Now find : $$(\alpha - \beta)^2$$ = $$\alpha^2 + \beta^2 - 2k$$ = $$49 - 4k$$ . So $$\alpha + \beta = 7$$ and $$\alpha - \beta = \sqrt{49-4k}$$

15. myko Group Title

2 equations 3 unknowns.....

16. mathslover Group Title

$$2\alpha = -7 + \sqrt{49-4k}$$ $$2\alpha + 7 = \sqrt{49-4k}$$ $$(2\alpha + 7)^2 = 49-4k$$ $$4\alpha^2 + 49 + 28 \alpha = 49 -4k$$ $$4\alpha^2 + \cancel{49} + 28\alpha = \cancel{49} - 4k$$ $$4\alpha (\alpha + 7) = -4k$$ $$k = \cfrac{\cancel{4}^{-1}\alpha(\alpha+7)}{\cancel{4}^1}$$ $$k = -\alpha^2 -7\alpha$$

17. mathslover Group Title

@myko yes but after simplifying we get much better answer for k in terms of only one root $$\alpha$$ .

18. hartnn Group Title

you could have got that by using : k= -x^2-7x since, alpha (or beta) is a root, x= alpha k= -alpha^2-7*alpha.

19. mathslover Group Title

I just presented my method . Though, there may be some arguments like its longer method or something else . But I just wanted to let the asker know about this method also. There may be many other methods to arrive the same answer that I given above, but if all the methods are known to a person, then it is better than knowing only 1 method , as per my opinion. Oh yeah, right Hartnn.

20. myko Group Title

I don't see why it is better or worse... It is exactly same amount of info

21. hartnn Group Title

agreed @mathslover

22. mathslover Group Title

Thank you Hartnn .