Quadratic Equations

Since $p%2Cq%2Cr$ are in A.P., $q%3D%5Cfrac%7Bp%2Br%7D%7B2%7D.$ Step 1: For the quadratic equation $px%5E2%2Bqx%2Br%3D0$ to have all real roots, its discriminant must be non-negative: $q%5E2-4pr%20%5Cge%200.$
Step 2: Substituting $q%3D%5Cfrac%7Bp%2Br%7D%7B2%7D$ , $%5Cleft(%5Cfrac%7Bp%2Br%7D%7B2%7D%5Cright)%5E2%20-%204pr%20%5Cge%200.$ $%5CRightarrow%20%5Cfrac%7Bp%5E2%2B2pr%2Br%5E2-16pr%7D%7B4%7D%20%5Cge%200.$ $%5CRightarrow%20p%5E2-14pr%2Br%5E2%20%5Cge%200.$
Step 3: Dividing throughout by $pr%3E0$ , $%5Cfrac%7Bp%7D%7Br%7D%2B%5Cfrac%7Br%7D%7Bp%7D%20%5Cge%2014.$ Let $x%3D%5Cfrac%7Bp%7D%7Br%7D%3E0$ . Then, $x%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cge%2014.$
Step 4: Solving, $x%5E2-14x%2B1%20%5Cge%200%20%5CRightarrow%20x%20%5Cge%207%2B4%5Csqrt%7B3%7D%20%5Cquad%20%5Ctext%7Bor%7D%20%5Cquad%20x%20%5Cle%207-4%5Csqrt%7B3%7D.$ Hence, $%5Cleft%7C%5Cfrac%7Bp%7D%7Br%7D-7%5Cright%7C%5Cge%204%5Csqrt%7B3%7D.$

Step 1: For a quadratic equation:

ax%5E2%2Bbx%2Bc%3D0

If the roots are

%5Calpha

and

%5Cbeta

, then:

%5Calpha%2B%5Cbeta%3D-%5Cfrac%7Bb%7D%7Ba%7D%2C%20%5Cqquad%20%5Calpha%5Cbeta%3D%5Cfrac%7Bc%7D%7Ba%7D

Step 2: Given equation:

4x%5E2%2B2x-1%3D0

Here,

a%3D4%2C%5Cquad%20b%3D2%2C%5Cquad%20c%3D-1

Step 3: Sum of roots:

%5Calpha%2B%5Cbeta%3D-%5Cfrac%7B2%7D%7B4%7D%3D-%5Cfrac%7B1%7D%7B2%7D

Step 4: Express the other root

%5Cbeta

in terms of

%5Calpha

%5Cbeta%3D-%5Cfrac%7B1%7D%7B2%7D-%5Calpha

Step 5: Rewrite

%5Cbeta

%5Cbeta%3D-%5Cfrac%7B1%2B2%5Calpha%7D%7B2%7D%3D-2%5Calpha-1%20%5Cquad%20(%5Ctext%7Bequivalent%20form%7D)

Step 6: Hence, the other root is:

%5Cboxed%7B-2%5Calpha-1%7D

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Quadratic Equations

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs