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Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: 16 x^{2}+24 x+1
CBSE | Class 10 | Exercise 10C | Maths | Quadratic Equations | RS Aggarwal
Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: 16 x^{2}+24 x+1

Given:

\begin{array}{l} 16 x^{2}+24 x+1 \\ \Rightarrow 16 x^{2}-24 x-1=0 \end{array}

On comparing it with a x^{2}+b x+x=0, we get; a=16, b=-24 and c=-1

Discriminant D is given by:

\begin{array}{l} D=\left(b^{2}-4 a c\right) \\ =(-24)^{2}-4 \times 16 \times(-1) \end{array}

\begin{array}{l} =576+(64) \\ =640>0 \end{array}

Hence, the roots of the equation are real.

Roots \alpha and \beta are given by:

\begin{array}{l} \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-24)+\sqrt{640}}{2 \times 16}=\frac{24+8 \sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10})}{4} \\ \beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-24)-\sqrt{640}}{2 \times 16}=\frac{24-8 \sqrt{10}}{32}=\frac{8(3-\sqrt{10})}{32}=\frac{(3-\sqrt{10})}{4} \end{array}

Thus, the roots of the equation are \frac{(3+\sqrt{10})}{4} and \frac{(3-\sqrt{10})}{4}.

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