Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the following equation for x. {:[sqrt(5x-4)=x-2],[x=]:}

Solve the following equation for x x .\newline5x−4=x−2x= \begin{array}{l} \sqrt{5 x-4}=x-2 \\ x= \end{array}

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Domain and range of square root functions: equationsright chevron icon

Full solution

Q. Solve the following equation for x x .\newline5x−4=x−2x= \begin{array}{l} \sqrt{5 x-4}=x-2 \\ x= \end{array}
  1. Isolating xx: We are given the equation 5x−4=x−2\sqrt{5x-4} = x-2. To solve for xx, we need to isolate xx on one side of the equation.
  2. Squaring both sides: First, we square both sides of the equation to eliminate the square root. This gives us (5x−4)2=(x−2)2(\sqrt{5x-4})^2 = (x-2)^2.
  3. Expanding the equation: Squaring both sides results in 5x−4=(x−2)25x - 4 = (x - 2)^2. Now we need to expand the right side of the equation.
  4. Bringing terms to one side: Expanding (x−2)2(x - 2)^2 gives us x2−4x+4x^2 - 4x + 4. So our equation now is 5x−4=x2−4x+45x - 4 = x^2 - 4x + 4.
  5. Factoring the quadratic equation: To solve for x, we need to bring all terms to one side to set the equation to zero. We subtract 55x and add 44 to both sides, resulting in 00 = x^22 - 44x - 55x + 44 + 44.
  6. Setting factors equal to zero: Combining like terms, we get 0=x2−9x+80 = x^2 - 9x + 8. This is a quadratic equation in standard form.
  7. Solving for x: To solve the quadratic equation, we can factor it. The factors of 88 that add up to −9-9 are −1-1 and −8-8. So we can write the equation as 0=(x−1)(x−8)0 = (x - 1)(x - 8).
  8. Checking solutions: Setting each factor equal to zero gives us two possible solutions for x: x−1=0x - 1 = 0 or x−8=0x - 8 = 0.
  9. Checking solutions: Setting each factor equal to zero gives us two possible solutions for x: x−1=0x - 1 = 0 or x−8=0x - 8 = 0. Solving x−1=0x - 1 = 0 gives us x=1x = 1. Solving x−8=0x - 8 = 0 gives us x=8x = 8. So we have two potential solutions: x=1x = 1 and x=8x = 8.
  10. Checking solutions: Setting each factor equal to zero gives us two possible solutions for x: x−1=0x - 1 = 0 or x−8=0x - 8 = 0. Solving x−1=0x - 1 = 0 gives us x=1x = 1. Solving x−8=0x - 8 = 0 gives us x=8x = 8. So we have two potential solutions: x=1x = 1 and x=8x = 8. We must check these solutions in the original equation to ensure they do not result in taking the square root of a negative number, as that would be invalid in the real number system.
  11. Checking solutions: Setting each factor equal to zero gives us two possible solutions for x: x−1=0x - 1 = 0 or x−8=0x - 8 = 0. Solving x−1=0x - 1 = 0 gives us x=1x = 1. Solving x−8=0x - 8 = 0 gives us x=8x = 8. So we have two potential solutions: x=1x = 1 and x=8x = 8. We must check these solutions in the original equation to ensure they do not result in taking the square root of a negative number, as that would be invalid in the real number system. Checking x=1x = 1 in the original equation extsqrt(5x−4)=x−2 ext{sqrt}(5x-4) = x-2, we get x−8=0x - 8 = 000, which simplifies to x−8=0x - 8 = 011. Since x−8=0x - 8 = 022, this solution does not satisfy the original equation.
  12. Checking solutions: Setting each factor equal to zero gives us two possible solutions for x: x−1=0x - 1 = 0 or x−8=0x - 8 = 0. Solving x−1=0x - 1 = 0 gives us x=1x = 1. Solving x−8=0x - 8 = 0 gives us x=8x = 8. So we have two potential solutions: x=1x = 1 and x=8x = 8. We must check these solutions in the original equation to ensure they do not result in taking the square root of a negative number, as that would be invalid in the real number system. Checking x=1x = 1 in the original equation extsqrt(5x−4)=x−2 ext{sqrt}(5x-4) = x-2, we get x−8=0x - 8 = 000, which simplifies to x−8=0x - 8 = 011. Since x−8=0x - 8 = 022, this solution does not satisfy the original equation. Checking x=8x = 8 in the original equation extsqrt(5x−4)=x−2 ext{sqrt}(5x-4) = x-2, we get x−8=0x - 8 = 055, which simplifies to x−8=0x - 8 = 066. Since x−8=0x - 8 = 066, this solution satisfies the original equation.

More problems from Domain and range of square root functions: equations

Question
Solve the following equation for y y .\newline2y+5=15−2yy= \begin{array}{l} 2 y+5=\sqrt{15-2 y} \\ y= \\ \end{array}
Get tutor helpright-arrow

Posted 10 months ago

Question
Solve the following equation for w w .\newline5−w4=w+8w= \begin{array}{l} \sqrt{5-\frac{w}{4}}=\sqrt{w+8} \\ w= \end{array}
Get tutor helpright-arrow

Posted 10 months ago

Question
Solve the following equation for z z .\newline4z+9=z+1z=□ \begin{array}{l} \quad \sqrt{4 z+9}=z+1 \\ z=\square \end{array}
Get tutor helpright-arrow

Posted 10 months ago

Question
Solve the following equation for y y .\newline2y−3=3y2−10xy=□ \begin{array}{l} 2 y-3=\sqrt{3 y^{2}-10 x} \\ y=\square \end{array}
Get tutor helpright-arrow

Posted 10 months ago

Question
lim⁡x→32x−5−1x−3= \lim _{x \rightarrow 3} \frac{\sqrt{2 x-5}-1}{x-3}=
Get tutor helpright-arrow

Posted 9 months ago

Question
lim⁡x→−4x+43x+13−1= \lim _{x \rightarrow-4} \frac{x+4}{\sqrt{3 x+13}-1}=
Get tutor helpright-arrow

Posted 9 months ago

Question
lim⁡x→−1x+1x+5−2= \lim _{x \rightarrow-1} \frac{x+1}{\sqrt{x+5}-2}=
Get tutor helpright-arrow

Posted 9 months ago

Question
Let f(x)=x+9 f(x)=\sqrt{x+9} and let c c be the number that satisfies the Mean Value Theorem for f f on the interval [0,16] [0,16] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) 1616\newline(B) 4343\newline(C) 5555\newline(D) 77
Get tutor helpright-arrow

Posted 9 months ago

Question
Let g(x)=5x−1 g(x)=\sqrt{5 x-1} and let c c be the number that satisfies the Mean Value Theorem for g g on the interval [1,10] [1,10] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) 22.2525\newline(B) 44.2525\newline(C) 66.55\newline(D) 88
Get tutor helpright-arrow

Posted 9 months ago

Question
You pick a card at random, put it back, and then pick another card at random.\newline44\newline55\newline66\newline77\newlineWhat is the probability of picking a number greater than 55 and then picking a 44 ?\newlineWrite your answer as a percentage.
Get tutor helpright-arrow

Posted 1 month ago

View all (19)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant