Chapter 8, Problem 1CR

To determine

To calculate: The real solutions of equation $3x^2+1=4x$, if any

Answer to Problem 1CR

Solution:

The real solutions of equation $3x^2+1=4x$ are $x=1,x=\frac{1}{3}$

Explanation of Solution

Given information:

The equation $3x^2+1=4x$

Formula used:

The standard form of quadratic equation is $a{x}^2+bx+c=0$

If $b^2-4ac>0$, then the equation has two different solutions.

If $b^2-4ac=0$, then the equation has repeated roots.

If $b^2-4ac<0$, then the equation has no real solution.

Solution of quadratic equation is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Convert the equation $3x^2+1=4x$ in standard form of quadratic equation by subtracting $4x$ from both sides.

  $\Rightarrow 3x^2-4x+1=0$

Compare the above equation with standard equation to find the values of $a,b$ and $c$

Therefore, $a=3,b=-4$ and $c=1$

Substitute the values of $a,b$ and $c$ in $b^2-4ac$

Therefore, $b^2-4ac={(-4)}^2-4(3)(1)=16-12=4$

  $b^2-4ac=4>0$

Therefore, the equation $3x^2+1=4x$ has two distinct real solutions.

Substitute values of $a,b$ and $c$ in quadratic formula.

  $\Rightarrow x=\frac{4\pm \sqrt{4}}{2\cdot 3}=\frac{4\pm 2}{6}$

  $\Rightarrow x=\frac{4+2}{6},x=\frac{4-2}{6}$

  $\Rightarrow x=1,x=\frac{2}{6}$

  $\Rightarrow x=1,x=\frac{1}{3}$ .

Therefore, the equation $3x^2+1=4x$ has real roots $x=1,x=\frac{1}{3}$