Test: Introduction to Quadratic Equations - Grade 9 MCQ

If the coefficient of x² is zero , then the equation is not a quadratic equation , its a linear equation. So its necessary condition for the quadratic equation. 

A quadratic equation in variable x is of the form ax²+ bx + c = 0, where a, b, c are real numbers a ≠ o, because if a=0 then the equation becomes a linear equation.
If we can factorise ax² + bx + c, a ≠ 0 into product of two linear factors then roots can be found by equating each factor to zero because if two factors are in multiplication and equal to zero then either of the factor is zero.
A real number R is said to be a root of the quadratic equation ax² + bx + c = 0 if a(R)² + bR + c = 0. , root means that the value gives answer equal to zero.
So all are correct.

The expression aⁿxⁿ+aⁿ⁻¹xⁿ⁻¹+⋯+a¹x+a^0​ is a polynomial where the highest power of the variable x is n.

This is a polynomial of degree n, because the degree of a polynomial is determined by the highest exponent of the variable x.

Thus, the correct answer is: d) polynomial of degree n.

Explanation:

- Let the two positive numbers be x and y.
- According to the given conditions:
- y = x + 5 (The two positive numbers differ by 5)
- (x + y)² = 169 (Square of their sum is 169)

- Substituting y = x + 5 in the second condition:
- (x + x + 5)² = 169
- (2x + 5)² = 169
- 4x² + 20x + 25 = 169
- 4x² + 20x - 144 = 0
- x² + 5x - 36 = 0
- (x + 9)(x - 4) = 0

- Therefore, the possible values for x are 4 and -9. Since the numbers are positive, x = 4.
- Substituting x = 4 in y = x + 5, we get y = 9.

- Therefore, the two positive numbers are 4 and 9, which is option C.

4 is the solution , this means that if we put x=4 we get 0. So putting x=4 in the equation x²+3x+k=0 we get 4²+3*4+k=0
16+12+k=0 ⇒ k=-28

Let, p(x) = x²-10x+k
since, 8 is the root of p(x)
∴ p(8) = 0
8²-10(8)+k = 0
64-80+k = 0
-16+k = 0
⇒ k = 16 .

option (a) is not Quadratic as it contain 1/x term



 

Conclusion:

Since all the given options b, c, and d simplify to quadratic equations, it seems that option a might be the one that is not a quadratic equation, although the exact expression is missing.

We have 5z²=3z
5z²-3z=0
z(5z-3)=0
So either z=0
Or 5z-3 =0  = z=⅗. So there are two solutions

sum of root = -b/a
for (3) option sum of roots will be -(3)/(-1) = 3

Option (B) and (D) , both are the correct answers.  We have x(x + 1) + 8 = (x + 2) (x – 2)
=x² + x + 8 = x² - 4
= x = -12, which is not a quadratic equation
Also, in (B) (x + 2)² = x³ – 4
=x² +4x + 4=x³ - 4, which is a cubic equation

Root of the equation means that the value when substituted in the equation gives zero as answer.
x² - kx + 5 = 0
Putting x = 1
1*1 -k + 5 = 0
-k+6=0
k=6

We have 
Taking LCM,

Multiplying LHS and RHS by x
x²+1=4x
x²-4x+1=0
Which is the required equation.

Given:

• x = 1 is a common root of the equations:
  • x² + ax - 3 = 0
  • bx² - 7x + 2 = 0

To find:

• The value of ab.

Solution:

Since x = 1 is a root of both equations, we can substitute x = 1 in both equations.

Equation 1:

• 1² + a(1) - 3 = 0
• 1 + a - 3 = 0
• a = 2

Equation 2:

• b(1)² - 7(1) + 2 = 0
• b - 7 + 2 = 0
• b = 5

Therefore, ab = 2 * 5 = 10

So, the value of ab is 10.

x² - √3x - x + √3 = 0
Taking common x from first two terms
x (x - √3) - 1 (x - √3) = 0
(x - √3) (x - 1) = 0
So, either x = √3 or x = 1

For ax² + bx + c = 0 to be a quadratic equation a0because if a is zero then ax² = 0 So we are left with only bx + c = 0 which is a linear equation in one variable. So to be a quadratic equation ax² cannot be zero.

Zeros of the polynomial means the value of variable such that the equation is equal to zero.Roots of the equation means the value of the variable for which LHS=RHS which basically means that the equation is equal to zero. Hence Zeros and roots are one and the same thing.