Friday

March 27, 2015

March 27, 2015

Posted by **kim** on Sunday, December 3, 2006 at 6:13pm.

Prove that the roots of

ax^2 + (a + b)x+b are real for all values of k

note the "x"s aren't multiplication signs.

a x^2 + bx + c has the discriminant of

D = b^2 - 4ac.

If D is nonnegative then the function has real roots.

In this case you have

D = (a + b)^2 - 4 a b = (a-b)^2

which is larger than or equal to zero because it is a square.

**Answer this Question**

**Related Questions**

MATH,HELP - Can someone show me how to even do this problem. Find all positive ...

Precalculus - "Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the ...

maths - if a, b,c are real numbers and not all equal, prove that the equation (c...

maths Please help!!!!! - if a, b,c are real numbers and not all equal, prove ...

Math (Algebra) - For each degree 17 polynomial f with real coefficients, let sf ...

math - If the roots of the eqn x²-2cx+ab=0 are real and unequal, then prove that...

Algebra II - Which describes the number and type of roots of the equation x^2 -...

maths2 - Use the discriminant to determine the number of real roots the equation...

mathematics - Use the discriminant to determine the number of real roots the ...

Algebra 2/Trigonometry - How do I solve f(x)=x^3-6x^2+16x-96 using Descartes' ...