SAT MATH QUESTIONS ON QUADRATIC EQUATIONS

Question 1 :

If x² + 4x = 45 and x > 0, what is the value of (x + 2)?

Answer :

x² + 4x = 45

Subtract 45 from both sides.

x² + 4x = 45

Factor and solve for x.

(x + 9)(x - 5) = 0

x + 9 = 0  or  x - 5 = 0

x = -9  or x = 5

Since x > 0, k = 5.

x + 2 = 5 + 2

x + 2 = 7

Question 2 :

Find the sum and product of the solutions to the following quadratic equation.

2x² + 6x + 2 = 0

Answer :

2x² + 6x + 2 = 0

Divide both sides by 2.

x² + 3x + 1 = 0

Comparing

x² + 3x + 1 = 0

and 

ax² + bx + c = 0

we get

a = 1, b = 3 and c = 1

Therefore, 

sum of the roots  = -b/a

= -3/1

= -3

product of the roots = c/a

= 1/1

= 1

Question 3 :

x² - 5x + c = 0

In the quadratic equation above, c is a constant. In the equation has two solutions for x, one of which is -3, what is the value of the other solution?

Answer :

One of the roots of the quadratic equation is given, that is -3. Let k be the other root of the equation.

Comparing

x² - 5x + c = 0

and 

ax² + bx + c = 0

we get

a = 1, b = -5 and c = c

Therefore, 

sum of the roots = -b/a

= -(-5)/1

= 5

k + (-3) = 5

k - 3 = 5

Add 3 to both sides.

k = 8

So, the other solution is 8.

Question 4 :

If (x + 3)(x - 3) = 91, what is the value of x?

Answer :

(x + 3)(x - 3) = 91

x² - 3x + 3x - 9 = 91

x² - 9 = 91

Add 9 to both sides.

x² = 100

Take square root on both sides.

√x² = √100

 x = ±10

Question 5 :

(x - a)(x - a - b) = 0

In the quadratic equation above, a and b are constants greater than zero. if 8 and 3 are two solutions to the equation, what is the value of b.

Answer :

(x - a)(x - a - b) = 0

x - a = 0  or  x - a - b = 0

Given : 8 and 3 are two solutions (values of x) to the equation.

Assume a = 8 and a + b = 3.

Substitute a = 8 in a + b = 3.

8 + b = 3

Subtract 8 from both sides.

b = -5

Since b is a constant greater than zero, b = -5 can not be accepted. 

Assume a = 3 and a + b = 8.

Substitute a = 3 in a + b = 8.

3 + b = 8

Subtract 3 from both sides.

b = 5

In b = 5, the value of b is greater than zero. So, the value of b is 5.

Question 6 :

2x² - 10x + k = 0

The quadratic equation above, in which k is a constant has two solutions m and n. If m > n and m - n = 9, what is the value of n?

Answer :

Comparing

2x² - 10x + k = 0

and 

ax² + bx + c = 0

we get

a = 2, b = -10 and c = k

Sum of the roots = -b/a

= -(-10)/2

= 5

Given : The two solutions of the quadratic equation are m and n.

m + n = 5 ----(1)

Given : m - n = 9.

m - n = 9 ----(2)

Add (1) and (2) to eliminate n and solve for m.

2m = 14

Divide both sides by 2.

m = 7

Substitute m = 7 in (1).

7 + n = 5

Subtract 7 from both sides.

n = -2

Question 7 :

9x² + kx + 4 = 0

The quadratic equation above has two equal real roots, find the value of k.

Answer :

Comparing

9x² + kx + 1 = 0

and 

ax² + bx + c = 0

we get

a = 9, b = k and c = 1

If the quadratic equation in the form ax² + bx + c = 0 has two real roots, then the value of the discriminant b² - 4ac is zero.

b² - 4ac = 0

Substitute b = k, a = 9 and c = 1.

k² - 4(9)(1) = 0

k² - 36 = 0

Add 36 to both sides.

k² = 36

Take square root on both sides.

√k² = √36

 k = ±6

Question 8 :

kx² + 8x + 4 = 0

The quadratic equation above has two real roots, find the maximum possible value of k.

Answer :

Comparing

kx² + 8x + 4 = 0

and 

ax² + bx + c = 0

we get

a = k, b = 8 and c = 4

If the quadratic equation in the form ax² + bx + c = 0 has two real roots, then the value of the discriminant b² - 4ac is greater than or equal to zero.

b² - 4ac ≥ 0

Substitute b = 8, a = k and c = 4.

8² - 4(k)(4) ≥ 0

64 - 16k ≥ 0

Subtract 64 from both sides.

-16k ≥ -64

Divide both sides by -16.

k ≤ 4

k can be equal to 4 or less than 4.

Therefore, the maximum possible value for k is 4.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.