Question

Given that the quadratic equation x^2-2x-5=0 has 2 different roots, and a second quadratic equation has 2 roots, each of which is 2 less than the corresponding root of the given quadratic equation. If the second quadratic equation is x^2+ax+b=0, find the value of a and b.

Answer 1

can u hurry please :( im in a hurry

Answer 2

Step 1. solve using the quadratic formula getting the roots of the given quadratic equation. You should get:\[x _{1}=1+\sqrt{6},x _{2}=1-\sqrt{6}\]

Answer 3

Step 2. Since the problem requires that the roots of the new equation being 2 less than the roots found in step 1, subtract 2 from each root.\[1+\sqrt{6}-2=-1+\sqrt{6}\]\[1-\sqrt{6}-2=-1-\sqrt{6}\] Now convert these roots to factors of the new equation:\[(x+1-\sqrt{6)}(x+1+\sqrt{6})\] Multiply the factors to get the new equation.\[x ^{2}+2x-5\]

Answer 4

Now locate the coefficients that represent A and B in the standard form of a quadratic:
\[Ax ^{2}+Bx+C\]A=1
B=2