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AshyBoiii · Jun 14, 2012, 03:15 AM

(1) Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.
(2) prove that if the roots of ax^2 +bx+c = 0 are real so also are the roots of acx^2+(2ac-b^2)x +ac=0
(3)for that values of m is it possible to express 2x^2+3y^2+7xy+4x+my+z as the product of two factors linear in x and y .

ebaines · Jun 14, 2012, 06:31 AM

Originally Posted by AshyBoiii

(1) Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.

Are you sure that there is no factor of 2 in front of some more of the terms? If this was x^2 - 2(a+b+c)x+a^2+b^2+c^2+2bc -2ca-2ab it would be pretty easy.

Originally Posted by AshyBoiii

(2) prove that if the roots of ax^2 +bx+c = 0 are real so also are the roots of acx^2+(2ac-b^2)x +ac=0

If the roots of a quadratic are real, from the quadratic equation that means that b^2>4ac. Apply this to that second equation and see what you get.

Originally Posted by AshyBoiii

(3)for that values of m is it possible to express 2x^2+3y^2+7xy+4x+my+z as the product of two factors linear in x and y .

For this one I think they're looking for you to factor that equation into the form
(ax+by+c)(dx+ey+f). It takes a bit of playing around to figure out the values of the coefficients but you should be able to get it.

AshyBoiii · Jun 14, 2012, 07:42 AM

1) Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.
here there is no 2 as you say sir, and I am not good in maths. So if you can help me to these sums and teach me step by step it would be a great help sir. Thanks

ebaines · Jun 14, 2012, 02:37 PM

You can start by "completing the square" for the x^2 -(a+b+c)x part; do you know how to do that? (If not, see the explanation below*.) It will leave you with something that has the form: (x-M)^2 +N^2 where M and N are coefficients made up of a's, b's and c's. Because of the powers of 2 both parts of this expression may be positive or zero, but never negative. The hard part is doing the algebra for this.

*To complete the square for x^2 + Ax +B, you take the coefficient of the x term, divide it by 2, square that, and then add and subtract the result, giving you: x^2 + Ax + A^2/4 - A^2/4 + B. The first three terms are a square, so the original equation is equivalent to: (x+A/2)^2 - A^2/4 + B.

tanishq · Jun 19, 2012, 06:29 AM

((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca) verify the equation by figure

ebaines · Jun 19, 2012, 06:39 AM

Originally Posted by tanishq

((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca) verify the equation by figure

Please don't double-post questions. This question has its own thread here:
https://www.askmehelpdesk.com/mathematics/x-2-x-2-2ax-2-verify-equation-figure-672684.html