Question

How do you solve the equation ${{x}^{2}}-2=17$?

Answer 1

Hint: Now the given equation is a quadratic equation which can be written in the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$ . Now we know that the roots of the equation are given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence substituting the values in the formula and then simplifying we will get the roots of the given equation.

Complete step-by-step solution:
Now let us consider the given equation ${{x}^{2}}-2=17$ . We want to find the solution of the equation which means we want to find the value of x for which the equation holds.
Let us first write the equation in general form $a{{x}^{2}}+bx+c=0$
Transposing 17 on LHS we get the equation as ${{x}^{2}}-2-17=0$
On simplifying the equation we get, ${{x}^{2}}-19=0$ . Now the equation is in standard form.
Hence comparing the equation with the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$ we get, a = 1, b = 0 and c = -19.
Now we know that the roots of the equation are given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence substituting the values of a, b and c in the formula we get,
$\Rightarrow x=\dfrac{0\pm \sqrt{0-4\left( 1 \right)\left( -19 \right)}}{2\left( 1 \right)}$
Taking 4 out of the square root we get,
$\Rightarrow x=\dfrac{\pm 2\sqrt{19}}{2}=\pm \sqrt{19}$
Hence the roots of the equation are $x=\sqrt{19}$ or $x=-\sqrt{19}$ .
Hence $x=\sqrt{19}$ and $x=-\sqrt{19}$ are the solution of the given equation.

Note: Now we can also skip the formula method and solve this equation by an easy method. Since we have b = 0 we can directly take the square root of the equation. Now consider the example ${{x}^{2}}-19=0$. Taking 19 on RHS and then taking square roots we easily get the solution of the equation as $x=\pm 19$ . Similarly we can also use the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to expand the equation ${{x}^{2}}-19={{x}^{2}}-{{\left( \sqrt{19} \right)}^{2}}$ and solve the factors of the equation.