Quadratic Equations Test

Result

Quadratic Equations Test - 3

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Question 1
1 / -0

Only one of the roots of ax² + bx + c = 0, a ≠ 0, is zero, if

A
c = 0

B
c = 0, b ≠ 0

C
b = 0, c = 0

D
b ≠ 0, c ≠ 0

Solution

When one root is zero and the other is non-zero, then
a(0)² + b(0) + c = 0
i.e. c = 0

Now, b ≠ 0 because if b = 0, then both the roots will be equal to 0.
As the equation will become
ax²= 0

Or x = 0, 0 (not possible, as both the roots are zero)
i.e. c = 0 and b ≠ 0
Question 2
1 / -0

It is known that y² - 11y + 4k = 0 has two distinct roots. It is also known that one factor is (y - k) and k is greater than zero. What is the difference between the sum and products of the roots?

A
28

B
17

C
11

D
7

Solution

Since (y - k) is a factor of the quadratic equation, so substitute y = k in the equation.
So the equation y² - 11y + 4k = 0 transforms to k² - 11k + 4k = 0
Hence, k² = 7k and k = 7.

Now, substitute k = 7 in the equation y² - 11y + 4k = 0
The equation becomes y² - 11y + 28 = 0, the roots of which are 7 and 4.

Sum of the roots = 11;
Product of roots = 28
So, required difference = 28 - 11 = 17
Question 3
1 / -0

For what value of m will the equation (m + 1)x² + 2(m + 3)x + m + 8 = 0 have equal roots?

A
-1/2

B
1/3

C
-2

D
1

Solution

For equal roots, D = 0 or b² - 4ac = 0
4(m + 3)² - 4(m + 1)(m + 8) = 0
(m + 3)² - (m + 1)(m + 8) = 0
m² + 9 + 6m - m² - 9m - 8 = 0
-3m + 1 = 0
m = 1/3
Question 4
1 / -0

The values of a for which the expression (a² - 1)x² + 2 (a - 1)x + 2 is positive for any x are:

A
a ≥ 1

B
a ≤ 1

C
a > -3 or a ≤ 1

D
a < -3 or a ≥ 1

Solution

We know that the expression ax² + bx + c > 0 for all x, if a > 0 and b² < 4ac.
1(a² - 1)x² + 2(a - 1)x + 2 is positive for all x.

If a² - 1 > 0 and 4(a - 1)² - 8(a² - 1) < 0,
a² - 1 > 0 and -4(a - 1)(a + 3) < 0
⇒ a² - 1 > 0 and (a - 1)(a + 3) > 0

⇒ a < -1 or a > 1 and a < -3 or a > 1
⇒ a < -3 or a > 1

But at a = 1, given expression is always positive for any x
Therefore a < -3 or a ≥ 1