# SAT MATH QUESTIONS ON QUADRATICS

Question 1 :

The graph of a parabola in the xy-plane has x-intercepts 3/5 and -1/2. Which of the following could be the equation of the parabola?

A) y = (5x - 1)((2x + 3)

B) y = (5x + 1)(2x - 3)

C) y = (5x - 3)(2x + 1)

D) y = (5x + 3)(2x - 1)

To find x-intercepts of any curve, we have to plugin y = 0 into its equation and find the values of x.

In each of the answer choices, plugin y = 0 and find x-intercepts.

A) y = (5x - 1)((2x + 3) :

(5x - 1)((2x + 3) = 0

5x - 1 = 0  or  2x + 3 = 0

x = 1/5  or  x = -3/2

x-intercepts are 1/5 and -3/2.

B) y = (5x + 1)((2x - 3) :

(5x + 1)((2x - 3) = 0

5x + 1 = 0  or  2x - 3 = 0

x = -1/5  or  x = 3/2

x-intercepts are -1/5 and 3/2.

C) y = (5x - 3)(2x + 1) :

(5x - 3)((2x + 1) = 0

5x - 3 = 0  or  2x + 1 = 0

x = 3/5  or  x = -1/2

x-intercepts are 3/5 and -1/2.

D) y = (5x + 3)(2x - 1) :

(5x + 3)((2x - 1) = 0

5x + 3 = 0  or  2x - 1 = 0

x = -3/5  or  x = 1/2

x-intercepts are -3/5 and 1/2.

The correct answer choice is (C).

Question 2 :

If k2 + 4k = 45 and k > 0, what is the value of (k + 2)?

k2 + 4k = 45

Subtract 45 from both sides.

k2 + 4k - 45 = 0

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

k + 9 = 0  or  k - 5 = 0

k = -9  or k = 5

Since k > 0, k = 5.

k + 2 = 5 + 2

k + 2 = 7

Question 3 :

The function f is defined by f(x) = x2 + bx + c, where b and c are constants. If the graph of f has x-intercepts at -5 and 3, which of the following correctly gives the values of b and c?

A) b = -5 and c = 3

B) b = -3 and c = 5

C) b = -2 and c = -15

D) b = 2 and c = -15

Given : x-intercepts are -5 and 3.

That is, when f(x) = 0, x = -5 or x = 3.

 (-5)2 + b(-5) + c = 025 - 5b + c = 0-5b + c = -25 ----(1) 32 + b(3) + c = 09 + 3b + c = 03b + c = -9 ----(2)

(2) - (1) :

(3b + c) - (-5b + c) = -9 - (-25)

3b + c + 5b - c = -9 + 25

8b = 24

b = 3

Substitute b = 3 in (2).

3(3) + c = -9

9 + c = -9

c = -18

Question 4 :

What is the sum of the solutions to 2x2 - 6x + 2 = 0?

A) -3

B) -1

C) 1

D) 3

2x2 - 6x + 2 = 0

In the quadratic equation above, the coefficients of x2, x and constant are all divisible by 2. So, divide both sides of the equation by. to make the values smaller.

x2 - 3x + 1 = 0

Comparing x2 - 3x + 1 = 0 and ax2 + bx + c = 0,

a = 1, b = -3 and c = 1

Formula to find the sum of the solutions of a quadratic equation in standard form ax2 + bx + c = 0 :

= -b/a

Substitute a = 1 and b = -3.

= -(-3)/1

= 3

The correct answer choice is (D).

Question 5 :

x2 - 5x + c = 0

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

Given : For the given quadratic equation, one of the solutions is -3.

Let k be the other solution.

Comparing x2 - 5x + c = 0 and ax2 + bx + c = 0,

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

sum of the solutions = -b/a

-3 + k = -(-5)/1

-3 + k = 5

k = 8

The other solution is 8.

Question 6 :

y = x2 - 2x - 3

A parabola in the xy-plane is given by the equation above. Which of the following equivalent forms of the equation displays the coordinates of the vertex of parabola as constants or coefficients?

A) y = (x - 1)2 - 4

B) y = (x - 1)2 - 2

C) y = (x - 3)(x + 1)

D) y + 3 = x(x + 2)

The vertex form equation of a parabola y = a(x - h)2 + k, displays the coordinates of the vertex (h, k) as constants or coefficients.

Write the given equation of the parabola in vertex form using completing the square method.

y = x2 - 2x - 3

y = x2 - 2(x)(1) - 3

y = x2 - 2(x)(1) + 12 - 12 - 3

Using the algebraic identity, (a + b)2 = a2 + 2ab + b2,

y = (x - 1)2 - 12 - 3

y = (x - 1)2 - 1 - 3

y = (x - 1)2 - 4

The correct answer choice is open (A).

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