Exercises and Answers with Explanations on Quadratic Equations
Question 1
Solve the equation: x 2 − 5 x +6=0.
An sw er
x 1=2, x 2=3.
Explan ation
For the quadratic equation a x 2 +bx+ c=0 (a ≠ 0), here a=1, b=−5 , c=6. We can use the
factorization method. Factor the equation as (x − 2)( x −3)=0, then x − 2=0 or x − 3=0.
Thus, the solutions are x 1=2 and x 2=3.
Question 2
Given that one root of the equation x 2+ kx − 6=0 is 2, find the value of k and the other
root.
An sw er
k =1, and the other root is −3 .
Explan ation
Substitute x=2 into the equation x 2+ kx − 6=0, we get 22 +2 k −6=0, that is 4 +2 k −6=0
, 2 k −2=0. Solving this gives k =1. The original equation becomes x 2+ x −6=0 .
Factorizing it, we have (x +3)(x − 2)=0, so the other root is x=− 3.
Question 3
Solve the equation by completing the square: x 2+ 4 x −5=0.
An sw er
x 1=1, x 2=−5 .
Explan ation
First, add the square of half the coefficient of the linear term to both sides of the
equation. That is, x 2+ 4 x +4 −4 −5=0, which is ¿. Shifting terms, we get ¿. Taking
, square roots, we have x +2=± 3, so x +2=3 or x +2=− 3. Thus, the solutions are x 1=1
and x 2=−5 .
Question 4
If the quadratic equation about x , (m −1)x 2+ 2 x +m 2 − 1=0 , has a root of 0 , find the
value of m .
An sw er
m=− 1.
Explan ation
Substitute x=0 into the equation (m −1) x 2+ 2 x +m 2 − 1=0 , we get m2 − 1=0 , that is
(m+1)(m−1)=0. The solutions are m=1 or m=− 1. Since it is a quadratic equation, the
coefficient of the quadratic term m− 1≠ 0 , that is m≠ 1. Therefore, m=− 1.
Question 5
Solve the equation: 2 x2 −3 x − 2=0 .
An sw er
1
x 1=2, x 2=− .
2
Explan ation
Use the quadratic formula x= −b ± √ b −4 ac , where a=2, b=−3 , c=− 2.
2
2a
2 3 ± √ 25 3 ±5 . So, x = 3+5 =2 and x = 3− 5 =− 1 .
Δ=b − 4 ac =¿. Then x= = 1 2
2 ×2 4 4 4 2
Question 6
Given that the quadratic equation x 2 − 4 x +k =0 has two equal real roots, find the value
of k .
An sw er
Question 1
Solve the equation: x 2 − 5 x +6=0.
An sw er
x 1=2, x 2=3.
Explan ation
For the quadratic equation a x 2 +bx+ c=0 (a ≠ 0), here a=1, b=−5 , c=6. We can use the
factorization method. Factor the equation as (x − 2)( x −3)=0, then x − 2=0 or x − 3=0.
Thus, the solutions are x 1=2 and x 2=3.
Question 2
Given that one root of the equation x 2+ kx − 6=0 is 2, find the value of k and the other
root.
An sw er
k =1, and the other root is −3 .
Explan ation
Substitute x=2 into the equation x 2+ kx − 6=0, we get 22 +2 k −6=0, that is 4 +2 k −6=0
, 2 k −2=0. Solving this gives k =1. The original equation becomes x 2+ x −6=0 .
Factorizing it, we have (x +3)(x − 2)=0, so the other root is x=− 3.
Question 3
Solve the equation by completing the square: x 2+ 4 x −5=0.
An sw er
x 1=1, x 2=−5 .
Explan ation
First, add the square of half the coefficient of the linear term to both sides of the
equation. That is, x 2+ 4 x +4 −4 −5=0, which is ¿. Shifting terms, we get ¿. Taking
, square roots, we have x +2=± 3, so x +2=3 or x +2=− 3. Thus, the solutions are x 1=1
and x 2=−5 .
Question 4
If the quadratic equation about x , (m −1)x 2+ 2 x +m 2 − 1=0 , has a root of 0 , find the
value of m .
An sw er
m=− 1.
Explan ation
Substitute x=0 into the equation (m −1) x 2+ 2 x +m 2 − 1=0 , we get m2 − 1=0 , that is
(m+1)(m−1)=0. The solutions are m=1 or m=− 1. Since it is a quadratic equation, the
coefficient of the quadratic term m− 1≠ 0 , that is m≠ 1. Therefore, m=− 1.
Question 5
Solve the equation: 2 x2 −3 x − 2=0 .
An sw er
1
x 1=2, x 2=− .
2
Explan ation
Use the quadratic formula x= −b ± √ b −4 ac , where a=2, b=−3 , c=− 2.
2
2a
2 3 ± √ 25 3 ±5 . So, x = 3+5 =2 and x = 3− 5 =− 1 .
Δ=b − 4 ac =¿. Then x= = 1 2
2 ×2 4 4 4 2
Question 6
Given that the quadratic equation x 2 − 4 x +k =0 has two equal real roots, find the value
of k .
An sw er
