Solving Quadratic Equations Using Square Roots

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Exercises 1 Let's review how can we find the solutions to an equation of the form x2=d by using square roots. For this, we can take the square root of each side of the equation to isolate x. Notice that there are three possible cases according to the value of d.Case d>0. In this case, since d​​=-d​, the equation x2=d has two real solutions, x=±d​. Case d=0. In this case, since 0​=-0​, the equation x2=d has one real solution, x=0. Case d<0. In this case, since d​ is not a real number, the equation x2=d has no real solutions.With this in mind we can complete the exercise's sentence.The equation x2=d has two real solutions when d>0.
Exercises 2 We are given the equations shown below. ​x2=144​x2−144=0​ x2+146=2 ​  x2+2=146 ​​ We are asked to solve them using square roots. To do this, we should first isolate the term x2 if needed, and then take the square root of both sides. For example, the equation x2=144 is already isolated, so we can take the square root directly. x2=144LHS​=RHS​x=±144​Calculate rootx=±12 Notice that we have two possible solutions: x=12, and x=-12. This is because the square of a positive number is equal to the square of its opposite, so we need to consider both signs. We can proceed similarly with the rest of the equations.EquationIsolate x2Solutions x2=144x2=144x=12 and x=-12 x2−144=0x2=144x=12 and x=-12 x2+146=2x2=-144No real solutions x2+2=146x2=144x=12 and x=-12 Note that all of the equations are equivalent to x2=144, having the solutions x=12 and x=-12. There is only one exception to this rule — the equation x2+146=2, which is equivalent to the equation x2=-144 and has no real solutions. Therefore, this is the one that is different.
Exercises 3 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution. Notice that in our case d=25, which is positive, so there are two real solutions to the given equation. To find them by taking the square roots, we need to consider the positive and negative solutions. x2=25LHS​=RHS​x=±25​Calculate rootx=±5 We found that x=±5. Therefore, there are two solutions for the equation, x=5 and x=-5.
Exercises 4 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution.Notice that in our case d=-36, which is negative, so there are no real solutions to the given equation.
Exercises 5 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution.Notice that in our case d=-21, which is negative, so there are no real solutions to the given equation.
Exercises 6 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution. Notice that in our case d=400, which is positive, so there are two real solutions to the given equation. To find them by taking the square roots, we need to consider the positive and negative solutions. x2=400LHS​=RHS​x=±400​Calculate rootx=±20 We found that x=±20. Therefore, there are two solutions for the equation, x=20 and x=-20.
Exercises 7 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution.Notice that in our case d=0, so there is only one real solution to the given equation, x=0.
Exercises 8 Before we will try to solve the given equation, let's recall the number of real solutions to a quadratic equation x2=d, depending on the sign of d.When d>0, x2=d has two real solutions, x=±d​. When d=0, x2=d has one real solution, x=0. When d<0, x2=d has no real solution. Notice that in our case d=169, which is positive, so there are two real solutions to the given equation. To find them by taking the square roots, we need to consider the positive and negative solutions. x2=169LHS​=RHS​x=±169​Calculate rootx=±13 We found that x=±13. Thus, there are two solutions for the equation: x=13 and x=-13.
Exercises 9 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. x2−16=0LHS+16=RHS+16x2=16LHS​=RHS​x=±16​Calculate rootx=±4 We found that x=±4. Therefore, there are two solutions for the equation, x=4 and x=-4.
Exercises 10 To solve the given equation by taking square roots, we need to start by isolating a variable raised at the power of two. x2+6=0LHS−6=RHS−6x2=-6 Remember, that the square root of the real number cannot be negative. Therefore, the equation has no real solutions.
Exercises 11 To solve the given equation by taking square roots, we need to start by isolating a variable raised at the power of two. 3x2+12=0LHS−12=RHS−123x2=-12LHS/3=RHS/3x2=-4 Remember, that the square root of the real number cannot be negative. Therefore, the equation has no real solutions.
Exercises 12 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. x2−55=26LHS+55=RHS+55x2=81LHS​=RHS​x=±81​Calculate rootx=±9 We found that x=±9. Therefore, there are two solutions for the equation, x=9 and x=-9.
Exercises 13 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 2x2−98=0LHS+98=RHS+982x2=98LHS/2=RHS/2x2=49LHS​=RHS​x=±49​Calculate rootx=±7 We found that x=±7. Therefore, there are two solutions for the equation, x=7 and x=-7.
Exercises 14 To solve the given equation by taking square roots, we need to start by isolating x2 on one side of the equation. -x2+9=9LHS−9=RHS−9-x2=0LHS/(-1)=RHS/(-1)x2=0Calculate rootx=0 We found that x=0. Therefore, there is only one solution for the equation: x=0.
Exercises 15 To solve the given equation by taking square roots, we need to start by isolating x2 on one side of the equation. -3x2−5=-5LHS+5=RHS+5-3x2=0LHS/(-3)=RHS/(-3)x2=0Calculate rootx=0 We found that x=0. Therefore, there is only one solution for the equation: x=0.
Exercises 16 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 4x2−371=29LHS+371=RHS+3714x2=400LHS/4=RHS/4x2=100LHS​=RHS​x=±100​Calculate rootx=±10 We found that x=±10. Therefore, there are two solutions for the equation, x=10 and x=-10.
Exercises 17 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 4x2+10=11LHS−10=RHS−104x2=1LHS/4=RHS/4x2=41​LHS​=RHS​x=±41​​Calculate rootx=±21​ We found that x=±21​. Therefore, there are two solutions for the equation, x=21​ and x=-21​.
Exercises 18 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 9x2−35=14LHS+35=RHS+359x2=49LHS/9=RHS/9x2=949​LHS​=RHS​x=±949​​ba​​=b​a​​x=±9​49​​Calculate rootx=±37​ We found that x=±37​. Therefore, there are two solutions for the equation, x=37​ and x=-37​.
Exercises 19 To solve the given equation by taking square roots, we need to start by taking the square roots of both sides of the given equation. Then we will isolate the variable. (x+3)2=0LHS​=RHS​x+3=0LHS−3=RHS−3x=-3 We found that x=-3. Therefore, there is only one solution for the equation: x=-3.
Exercises 20 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. (x−1)2=4LHS​=RHS​x−1=±4​Calculate rootx−1=±2LHS+1=RHS+1x=1±2 We found that there are two solutions for the given equation.    x=1+2x=3​   x=1−2 x=-1​
Exercises 21 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. (2x−1)2=81LHS​=RHS​2x−1=±9LHS+1=RHS+12x=1±9LHS/2=RHS/2x=21​±29​ The solutions for this equation are x=21​±29​. Let's separate them by using the positive and negative signs.x=21​±29​ x=21​+29​x=21​−29​ x=210​x=-28​ x=5x=-4 We found that there are two solutions for the given equation: x=5 and x=-4.
Exercises 22 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. (4x+5)2=9LHS​=RHS​4x+5=±3LHS−5=RHS−54x=-5±3LHS/4=RHS/4x=-45​±43​ The solutions for this equation are x=-45​±43​. Let's separate them by using the positive and negative signs.x=-45​±43​ x=-45​+43​x=-45​−43​ x=-42​x=-48​ x=-21​x=-2 We found that there are two solutions for the given equation, x=-21​ and x=-2.
Exercises 23 To solve the given equation by taking square roots, we need to start from isolating a variable raised at the power of two. Then we will consider the positive and negative solutions. 9(x+1)2=16LHS/9=RHS/9(x+1)2=916​LHS​=RHS​x+1=±916​​ba​​=b​a​​x+1=±9​16​​Calculate quotientx+1=±34​LHS−1=RHS−1x=-1±34​ The solutions for this equation are x=-1±34​. Let's separate them by using the positive and negative signs.x=-1±34​ x=-1+34​x=-1−34​ x=-33​+34​x=-33​−34​ x=31​x=-37​ We found that there are two solutions for the given equation: x=31​ and x=-37​.
Exercises 24 To solve the given equation by taking square roots, we need to start from isolating a variable raised at the power of two. Then we will consider the positive and negative solutions. 4(x−2)2=25LHS/4=RHS/4(x−2)2=425​LHS​=RHS​x−2=±425​​ba​​=b​a​​x−2=±4​25​​Calculate rootx−2=±25​LHS+2=RHS+2x=2±25​ The solutions for this equation are x=2±25​. Let's separate them by using the positive and negative signs.x=2±25​ x=2+25​x=2−25​ x=24​+25​x=24​−25​ x=29​x=-21​ We found that there are two solutions for the given equation: x=29​ and x=-21​.
Exercises 25 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. x2+6=13LHS−6=RHS−6x2=7LHS​=RHS​x=±7​Use a calculatorx=±2.6457…x≈±2.65 We found that x≈±2.65. Therefore, there are two solutions for the equation, x≈2.65 and x≈-2.65.
Exercises 26 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. x2+11=24LHS−11=RHS−11x2=13LHS​=RHS​x=±13​Use a calculatorx=±3.6055…x≈±3.61 We found that x≈±3.61. Therefore, there are two solutions for the equation, x≈3.61 and x≈-3.61.
Exercises 27 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 2x2−9=11LHS+9=RHS+92x2=20LHS/2=RHS/2x2=10LHS​=RHS​x=±10​Use a calculatorx=±3.1622…x≈±3.16 We found that x≈±3.16. Therefore, there are two solutions for the equation, x≈3.16 and x≈-3.16.
Exercises 28 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 5x2+2=6LHS−2=RHS−25x2=4LHS/5=RHS/5x2=54​Write as a decimalx2=0.8LHS​=RHS​x=±0.8​Use a calculatorx=±0.8944…x≈±0.89 We found that x≈±0.89. Therefore, there are two solutions for the equation: x≈0.89 and x≈-0.89.
Exercises 29 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. -21=15−2x2LHS+2x2=RHS+2x22x2−21=15LHS+21=RHS+212x2=36LHS/2=RHS/2x2=18LHS​=RHS​x=±18​Use a calculatorx=±4.2426…x≈±4.24 We found that x≈±4.24. Therefore, there are two solutions for the equation: x≈4.24 and x≈-4.24.
Exercises 30 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 2=4x2−5LHS+5=RHS+57=4x2Rearrange equation4x2=7LHS/4=RHS/4x2=47​Write as a decimalx2=1.75LHS​=RHS​x=±1.75​Use a calculatorx=±1.3228…x≈±1.32 We found that x≈±1.32. Therefore, there are two solutions to the equation, x≈1.32 and x≈-1.32.
Exercises 31 We will solve the equation and then compare our solution with the given work. To solve the equation using square roots, we have to start by isolating x on the left-hand side of the equation. 2x2−33=39LHS+33=RHS+332x2=72LHS/2=RHS/2x2=36 So far our solution is identical with the given work. Our next step will be taking the square root of each side. Don't forget that when solving an equation this way we must consider both the positive and the negative square roots. x2=36LHS​=RHS​x=±36​Calculate rootx=±6 Therefore, the solutions to the equation are x=6 and x=-6. The given work is missing the negative square root of 36.
Exercises 32 Let's recall the formula for the volume of a rectangular prism. V=ℓwh​ Here ℓ is the length of the prism, w is the width of the prism, and h is its height — or in our case its depth. We know that V=72000 and h=24. We also know that the length ℓ is two times the width w, so ℓ=2w. We will substitute these values into the formula. By doing so we obtain an equation that can be solved for w. V=ℓwhSubstitute V=72000,h=24,ℓ=2w72000=2w(w)(24)Multiply72000=48w2LHS/48=RHS/481500=w2LHS​=RHS​±1500​=wRearrange equationw=±1500​ The solutions to the equation are 1500​ and -1500​. Since w represents the width of the pond, it cannot be negative. Therefore, w=1500​ and ℓ=21500​. We can use a calculator to approximate our results. wℓ​=1500​≈38.7=21500​≈77.5​ The width is 1500​≈38.7 inches and the length is 21500​≈77.5 inches.
Exercises 33 We are given a function that represents the height h of the dropped sunglasses after x seconds. h=-16x2+24​ The sunglasses hit the ground when h=0. Therefore, to determine how long it takes for the sunglasses to hit the ground we have to determine the value of x when h=0. Let's substitute h=0 into the equation and simplify! h=-16x2+24h=00=-16x2+24LHS+16x2=RHS+16x216x2=24 LHS/16=RHS/16 LHS/16=RHS/16x2=1624​ba​=b/8a/8​x2=23​Calculate quotient x2=1.5LHS​=RHS​x=±1.5​ The solutions to the equation are x=1.5​ and x=-1.5​. Since x represents time, it cannot be negative. Therefore, x=1.5​≈1.2. It takes about 1.2 seconds for the sunglasses to hit the ground.
Exercises 34 Let's try to solve the given equation using square roots. To do so, we have to isolate x2 on one side of the equation. x2+4=0⇔x2=-4​ Recall that a square of a real number cannot be negative. Therefore, the equation has no real solutions. Your cousin is correct.
Exercises 35 We will start by determining what the area of the inner square must be. Then we will use this information to find the side length x of the inner square.Area of the Inner Square We want the area of the inner square to be 25% of the total area of the rug. Let's determine the total area of the rug. We will use the formula for the area of a square. A=s2​ Here s is the side length of the square. In our case, s=6. Let's calculate the total area of the rug Ar​. Ar​=s2s=6Ar​=62Calculate powerAr​=36 Let Ai​ represent the area of the inner square. The area Ai​ must be equal to 25% of the area Ar​. Ai​=25%⋅Ar​=25%⋅36​ Let's simplify the obtained expression. Ai​=25%⋅36a%=100a​Ai​=10025​⋅36ca​⋅b=ca⋅b​Ai​=10025⋅36​ba​=b/25a/25​Ai​=436​Calculate quotientAi​=9 The area of the inner square must be 9 square feet.Side Length of the Inner Square We have determined that the area of the inner square must be 9 square feet. Since x represents the side length of the inner square, we have the following equation. x2=9​ We will solve the equation using square roots. However, we will only consider the positive square root, because a negative solution does not make sense in this context. x=9​=3​ The side length of the inner square is 3 feet.
Exercises 36
Exercises 37 Let's first review how to find the exact solution of a quadratic equation. For this, we need to isolate x. We can do this by first isolating x2. Then, we can calculate the square root of both sides to undo the power. For example, let's do this for x2−2. x2−2=0LHS+2=RHS+2x2=2LHS​=RHS​x=±2​ Notice that we have two possible solutions; one positive and one negative. For practical reasons, we may want to find an approximate solution instead. We can do this by using the calculator to find the square root value. Then, we can round to the desired number of decimal places according to our needs. We can see an example below.Solution 1Solution 2 x=2​x=-2​ x=1.414213…x=-1.414213… Round to the second decimal place x≈1.41x≈-1.42
Exercises 38
Exercises 39 Two functions intersect if their graphs have commons points. This happens when both functions have the same x- and y-values. Considering the given functions, we can use the Transitive Property of Equality to set an equation in terms of x. y=x2andy=9⇓x2=9​ To solve this equation, we will isolate the x-variable by taking square roots on both sides. x2=9LHS​=RHS​x=±9​Calculate rootx=±3 Notice that we have two solutions for the x-variable, x=-3 and x=3. This means that the functions will intersect at points with these x-coordinates. Since one of the given functions is y=9, we know that the y-coordinate of these points is 9.The functions y=x2 and y=9, intersect at the points (-3,9) and (3,9).
Exercises 40 Recall that the x-intercepts of the graph of a function f(x) are the solutions to the equation f(x)=0. Therefore, we can tell the number of solutions that a quadratic equation has by counting the number of times that its associated function intersects the x-axis. In this case, we have the graph shown below.As we can see, the graph has only one x-intercept. This means that the associated equation (x−1)2=0 has only one real solution.
Exercises 41 We are asked to solve the equation shown below. x2=1.44​ Since x2 is already isolated, we just need to calculate the square root of 1.44. Recall that 122=144. This is very similar to the calculation we would need to do. We can rewrite this equality by dividing it by 100 to find the square root of 1.44. 122=144 Rewrite equality LHS/100=RHS/100100122​=100144​Rewrite 100 as 102102122​=100144​bmam​=(ba​)m(1012​)2=100144​Calculate quotient ( 1.2)2=1.44 As we can see, 1.22=1.44. This means that 1.2 is the square root of 1.44, and we can use this result to solve our equation. x2=1.44LHS​=RHS​x=±1.44​Calculate rootx=±1.2 Therefore, the solutions are x=1.2 and x=-1.2.
Exercises 42 We are told that a quadratic equation in the form ax2+bx+c=0 can be rewritten as shown below. ax2+bx+c=0⇔(x+2ab​)2=4a2b2−4ac​​ To write the solutions of the equation, we will isolate the x-variable. We can start by taking square roots on both sides. Recall that we need to consider the positive and negative values. (x+2ab​)2=4a2b2−4ac​LHS​=RHS​x+2ab​=±4a2b2−4ac​​LHS−2ab​=RHS−2ab​x=-2ab​±4a2b2−4ac​​ba​​=b​a​​x=-2ab​±4a2​b2−4ac​​a⋅b​=a​⋅b​x=-2ab​±4​a2​b2−4ac​​Calculate rootx=-2ab​±2a2​b2−4ac​​ Note that in one of the denominators we have the expression a2​. Depending on the value of a, we can simplify this in two ways. a2​={-a,if a≥0-a,if a<0​​ However, since we are considering both the positive and negative signs for the term containing this expression, we can simplify a2​ as a. x=-2ab​±2a2​b2−4ac​​a2​=ax=-2ab​±2ab2−4ac​​Put minus sign in numeratorx=2a-b​±2ab2−4ac​​Add and subtract fractionsx=2a-b±b2−4ac​​ We have found that we can write the solutions of a quadratic equation in standard form as x=2a-b±b2−4ac​​.
Exercises 43 We want to find the x-coordinates of the points on the parabola which have y-coordinates of 9. Since these points are on the parabola, they must satisfy the equation of the parabola. Therefore, to find the x-coordinates we will substitute y=9 into the given equation and solve it for x. y=21​(x−2)2+1y=99=21​(x−2)2+1 To solve this equation we will rearrange it and isolate (x−2)2 on the left-hand side. 9=21​(x−2)2+1Rearrange equation21​(x−2)2+1=9LHS−1=RHS−121​(x−2)2=8LHS⋅2=RHS⋅2(x−2)2=16 Next we will take a square root of each side of the equation. Remember to calculate the positive and the negative square roots. (x−2)2=16LHS​=RHS​x−2=±16​Calculate rootx−2=±4LHS+2=RHS+2x=2±4 The solutions to the equation are x=2+4=6 and x=2−4=-2. Therefore, the points on the parabola which have y-coordinates of 9 are (6,9) and (-2,9).
Exercises 44
Exercises 45 We want to factor a perfect square trinomial. x2+8x+16​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?16=42 ✓ Is the middle term twice the product of 4 and x?8x=2⋅4⋅x ✓ As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is an addition sign in the middle. x2+8x+16⇔(x+4)2​
Exercises 46 We want to factor a perfect square trinomial. x2−4x+4​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?4=22 ✓ Is the middle term twice the product of 2 and x?4x=2⋅2⋅x ✓ As we can see, the answer to the three questions above is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is a subtraction sign in the middle. x2−4x+4⇔(x−2)2​
Exercises 47 We want to factor a perfect square trinomial. x2−14x+49​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?49=72 ✓ Is the middle term twice the product of 7 and x?14x=2⋅7⋅x ✓ As we can see, the answer to the three questions above is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is a subtraction sign in the middle. x2−14x+49⇔(x−7)2​
Exercises 48 We want to factor a perfect square trinomial. x2+18x+81​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?81=92 ✓ Is the middle term twice the product of 9 and x?18x=2⋅9⋅x ✓ As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is an addition sign in the middle. x2+18x+81⇔(x+9)2​
Exercises 49 We want to factor a perfect square trinomial. x2+12x+36​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?36=62 ✓ Is the middle term twice the product of 6 and x?12x=2⋅6⋅x ✓ As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is an addition sign in the middle. x2+12x+36⇔(x+6)2​
Exercises 50 We want to factor a perfect square trinomial. x2−22x+121​ How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.Is the first term a perfect square?x2=x2 ✓ Is the last term a perfect square?121=112 ✓ Is the middle term twice the product of 11 and x?22x=2⋅11⋅x ✓ As we can see, the answer to the three questions above is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is a subtraction sign in the middle. x2−22x+121⇔(x−11)2​