Solve $\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $=$ $1$

In this algebraic problem, an algebraic equation is defined in a variable $x$ as follows.

$\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $\,=\,$ $1$

The algebraic equation consists of a difference of two terms at left hand side and a number in right hand side. The left hand side expression is a difference of two terms. Each term is a square root of a quadratic expression. It indicates that we have to use the concept of quadratic equations for solving this algebraic equation.

Technique for solving the Algebraic equation

For solving the given algebraic equation $\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $\,=\,$ $1$, we have to release the quadratic expressions from the square roots. Mathematically, each quadratic expression can be released from the square root by taking square both sides of the equation. If we take this step, the equation becomes more complicated. Hence, we have to think for an alternative mathematical approach.

$\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $\,=\,$ $1$

Observe the first two terms in each quadratic expression. They both are the same. Hence, we can represent $5x^2-6x$ by a variable. For example, $y \,=\, 5x^2-6x$.

Now, we can write the given mathematical equation in terms of $y$ as follows.

$\implies$ $\sqrt{y+8}$ $-$ $\sqrt{y-7}$ $\,=\,$ $1$

Solve the equation in algebraic form

Take the square, both sides of the algebraic equation for eliminating the square root from the mathematical equation.

$\implies$ $\Big(\sqrt{y+8}$ $-$ $\sqrt{y-7}\Big)^2$ $=$ $(1)^2$

We can expand the square of the difference of the terms by the square of difference rule.

$\implies$ $\Big(\sqrt{y+8}\Big)^2$ $+$ $\Big(\sqrt{y-7}\Big)^2$ $-$ $2$ $\times$ $\sqrt{y+8}$ $\times$ $\sqrt{y-7}$ $\,=\,$ $1^2$

$\implies$ $y+8+y-7$ $-$ $2\sqrt{(y+8)(y-7)}$ $\,=\,$ $1$

$\implies$ $y+8+y-7-1$ $\,=\,$ $2\sqrt{(y+8)(y-7)}$

$\implies$ $y+8+y-8$ $\,=\,$ $2\sqrt{(y+8)(y-7)}$

$\implies$ $y+y+8-8$ $\,=\,$ $2\sqrt{(y+8)(y-7)}$

$\implies$ $2y+\cancel{8}-\cancel{8}$ $\,=\,$ $2\sqrt{(y+8)(y-7)}$

$\implies$ $2y$ $\,=\,$ $2\sqrt{(y+8)(y-7)}$

$\implies$ $\cancel{2}y$ $\,=\,$ $\cancel{2}\sqrt{(y+8)(y-7)}$

$\implies$ $y$ $\,=\,$ $\sqrt{(y+8)(y-7)}$

Take square both sides of the algebraic equation one more time for eliminating the square root completely.

$\implies$ $(y)^2$ $\,=\,$ $\Big(\sqrt{(y+8)(y-7)}\Big)^2$

$\implies$ $y^2$ $\,=\,$ $(y+8)(y-7)$

Now, multiply the binomials for expressing their product in algebraic form.

$\implies$ $y^2$ $\,=\,$ $y \times (y-7)$ $+$ $8 \times (y-7)$

$\implies$ $y^2$ $\,=\,$ $y \times y$ $-$ $y \times 7$ $+$ $8 \times y$ $-$ $8 \times 7$

$\implies$ $y^2$ $\,=\,$ $y^2$ $-$ $7y$ $+$ $8y$ $-$ $56$

$\implies$ $\cancel{y^2}$ $\,=\,$ $\cancel{y^2}$ $+$ $y$ $-$ $56$

$\implies$ $0$ $\,=\,$ $y$ $-$ $56$

$\implies$ $y$ $-$ $56$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $y$ $\,=\,$ $56$

Actually, we have taken that $y = 5x^2-6x$.

$\,\,\,\therefore\,\,\,\,\,\,$ $5x^2-6x$ $\,=\,$ $56$

$\implies$ $5x^2-6x-56$ $\,=\,$ $0$

Solve the quadratic equation

The algebraic equation $5x^2-6x-56$ $\,=\,$ $0$ is a quadratic equation. It can be solved for evaluating the variable $x$. The quadratic equation can be solved by the quadratic formula.

$x$ $\,=\,$ $\dfrac{-(-6)\pm\sqrt{(-6)^2-4 \times 5 \times (-56)}}{2 \times 5}$

$\implies$ $x$ $\,=\,$ $\dfrac{6\pm\sqrt{36+1120}}{10}$

$\implies$ $x$ $\,=\,$ $\dfrac{6\pm\sqrt{1156}}{10}$

$\implies$ $x$ $\,=\,$ $\dfrac{6\pm 34}{10}$

$\implies$ $x$ $\,=\,$ $\dfrac{6+34}{10}$ and $x$ $\,=\,$ $\dfrac{6-34}{10}$

$\implies$ $x$ $\,=\,$ $\dfrac{40}{10}$ and $x$ $\,=\,$ $\dfrac{-28}{10}$

$\implies$ $x$ $\,=\,$ $\dfrac{\cancel{40}}{\cancel{10}}$ and $x$ $\,=\,$ $\dfrac{-\cancel{28}}{\cancel{10}}$

$\implies$ $x$ $\,=\,$ $4$ and $x$ $\,=\,$ $\dfrac{-14}{5}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $4$ and $x$ $\,=\,$ $-\dfrac{14}{5}$

Verification

We have solved that $x$ $\,=\,$ $4$ and $x$ $\,=\,$ $-\dfrac{14}{5}$

$\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $=$ $1$

Now, let’s substitute each $x$ value in the left hand side expression to check whether its value is equal to one or not.

Let $x \,=\, 4$

$=\,\,\,$ $\sqrt{5(4)^2-6(4)+8}$ $-$ $\sqrt{5(4)^2-6(4)-7}$

$=\,\,\,$ $\sqrt{5 \times (4)^2-6 \times 4+8}$ $-$ $\sqrt{5 \times (4)^2-6 \times 4-7}$

$=\,\,\,$ $\sqrt{5 \times 16-24+8}$ $-$ $\sqrt{5 \times 16-24-7}$

$=\,\,\,$ $\sqrt{80-24+8}$ $-$ $\sqrt{80-24-7}$

$=\,\,\,$ $\sqrt{64}$ $-$ $\sqrt{49}$

$=\,\,\,$ $8-7$

$=\,\,\,$ $1$

Let $x \,=\, -\dfrac{14}{5}$

$=\,\,\,$ $\sqrt{5\bigg(-\dfrac{14}{5}\bigg)^2-6\bigg(-\dfrac{14}{5}\bigg)+8}$ $-$ $\sqrt{5\bigg(-\dfrac{14}{5}\bigg)^2-6\bigg(-\dfrac{14}{5}\bigg)-7}$

$=\,\,\,$ $\sqrt{5\bigg(\dfrac{196}{25}\bigg)-6\bigg(-\dfrac{14}{5}\bigg)+8}$ $-$ $\sqrt{5\bigg(\dfrac{196}{25}\bigg)-6\bigg(-\dfrac{14}{5}\bigg)-7}$

$=\,\,\,$ $\sqrt{5\bigg(\dfrac{196}{25}\bigg)+6\bigg(\dfrac{14}{5}\bigg)+8}$ $-$ $\sqrt{5\bigg(\dfrac{196}{25}\bigg)+6\bigg(\dfrac{14}{5}\bigg)-7}$

$=\,\,\,$ $\sqrt{\dfrac{5 \times 196}{25}+\dfrac{6 \times 14}{5}+8}$ $-$ $\sqrt{\dfrac{5 \times 196}{25}+\dfrac{6 \times 14}{5}-7}$

$=\,\,\,$ $\sqrt{\dfrac{\cancel{5} \times 196}{\cancel{25}}+\dfrac{6 \times 14}{5}+8}$ $-$ $\sqrt{\dfrac{\cancel{5} \times 196}{\cancel{25}}+\dfrac{6 \times 14}{5}-7}$

$=\,\,\,$ $\sqrt{\dfrac{1 \times 196}{5}+\dfrac{6 \times 14}{5}+8}$ $-$ $\sqrt{\dfrac{1 \times 196}{5}+\dfrac{6 \times 14}{5}-7}$

$=\,\,\,$ $\sqrt{\dfrac{196}{5}+\dfrac{84}{5}+8}$ $-$ $\sqrt{\dfrac{196}{5}+\dfrac{84}{5}-7}$

$=\,\,\,$ $\sqrt{\dfrac{196+84}{5}+8}$ $-$ $\sqrt{\dfrac{196+84}{5}+\dfrac{84}{5}-7}$

$=\,\,\,$ $\sqrt{\dfrac{280}{5}+8}$ $-$ $\sqrt{\dfrac{280}{5}-7}$

$=\,\,\,$ $\sqrt{56+8}$ $-$ $\sqrt{56-7}$

$=\,\,\,$ $\sqrt{64}$ $-$ $\sqrt{49}$

$=\,\,\,$ $8-7$

$=\,\,\,$ $1$

For $x = 4$ and $x \,=\, -\dfrac{14}{5}$, the value of algebraic expression $\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ is equal to $1$. Therefore, $x = 4$ and $x \,=\, -\dfrac{14}{5}$ are the solutions of the given algebraic equation.