Question

The number of the real roots of the equation ${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$ is

Answer 1

Step 1: State the given data and deduce two cases

The given equation is,

${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$ $...\left(\mathrm{i}\right)$

Now, as we know,

$\left|x\right|=x$ if $x\ge 0$

$\left|x\right|=-x$ if $x<0$

So, for the given equation,

$\left|x-5\right|=x-5$ if $x\ge 5$

$\left|x-5\right|=-\left(x-5\right)$ if $x<5$

Step 2: Simplify the given equation for $x\ge 5$

If $x\ge 5$, then the equation $\left(\mathrm{i}\right)$ becomes,

${\left(x+1\right)}^2+\left(x-5\right)=\frac{27}{4}$

$\Rightarrow$ $4{\left(x+1\right)}^2+4\left(x-5\right)=27$

$\Rightarrow$ $4x^2+8x+4+4x-20=27$ $\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

$\Rightarrow$ $4x^2+12x-43=0$

Step 3: Apply the quadratic formula

Now, by the quadratic formula, the root of a quadratic equation is expressed as,

$\begin{array}{cl}& x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow & x=\frac{-12\pm \sqrt{{\left(12\right)}^2-4\times 4\left(-43\right)}}{2\times 4}\\ \Rightarrow & x=\frac{-12\pm \sqrt{144+688}}{8}\\ \Rightarrow & x=\frac{-12\pm \sqrt{832}}{8}\\ \Rightarrow & x=\frac{-12\pm 8\sqrt{13}}{2}\end{array}$

i.e., there are no real roots of the equation $\left(\mathrm{i}\right)$ for $x\ge 5$.

Step 4: Simplify the given equation for $x<5$

If $x<5$, then the equation $\left(\mathrm{i}\right)$ becomes,

${\left(x+1\right)}^2-\left(x-5\right)=\frac{27}{4}$

$\Rightarrow$ $4{\left(x+1\right)}^2-4\left(x-5\right)=27$

$\Rightarrow$ $4x^2+8x+4-4x+20=27$ $\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

$\Rightarrow$ $4x^2+4x-3=0$

$\Rightarrow$ $4x^2+6x-2x-3=0$

$\Rightarrow$ $2x\left(2x+3\right)-1\left(2x+3\right)=0$

$\Rightarrow$ $\left(2x+3\right)\left(2x-1\right)=0$

So, $x=-\frac{3}{2}$ or $x=\frac{1}{2}$

Since both the values of $x$ are real.

Thus, the number of real roots $=2$

Hence, the number of the real roots of the equation ${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$ is $2$.