Question

21 points!!Determine the number of real solutions each quadratic equation has. Answer with either no, one, or two.
1.) y = 12x^2 - 9x + 4 has _ real solution(s)
2.) 10x + y = -x^2 + 2 has _ real solution(s)
3.) 4y - 7 = 5x^2 - x + 2 + 3y has _ real solution(s)
4.) y = (-x + 4)^2 has _ real solution(s)

Answer 1

Determine the number of real solutions each quadratic equation has. Answer with either no, one, or two.
1.) y = 12x^2 - 9x + 4 has _ real solution(s)
2.) 10x + y = -x^2 + 2 has _ real solution(s)
3.) 4y - 7 = 5x^2 - x + 2 + 3y has _ real solution(s)
4.) y = (-x + 4)^2 has _ real solution(s)

Answer 2

$y = 12\cdot x^{2}-9\cdot x + 4$ has no real solutions.

$10\cdot x + y = -x^{2}+2$  has two distinct real solutions.

$4\cdot y - 7 = 5\cdot x^{2}-x + 2 + 3\cdot y$ has no real solutions.

$y = (-x +4)^{2}$ has two equal real solutions.

In this exercise we need to determine the nature of solutions of given quadratic equations. According to the Fundamental Theorem of Algebra, a quadratic equation has two solutions, whose discriminant is:

$\delta = B^{2} - 4\cdot A\cdot C$ (1)

Where:

• $A$ - Second order coefficient.
• $B$ - First order coefficient.
• $C$ - Zero order coefficient.

The value of the discriminant tells us the nature of solutions of a given quadratic equation:

1) If $\delta > 0$, then the equation has two distinct real solutions.

2) If $\delta = 0$, then the equation has two equal real solutions.

3) If $\delta < 0$, then the equation has two conjugated complex solutions.

Now we proceed to check each expression:

1) $y = 12\cdot x^{2}-9\cdot x + 4$ ($A = 12, B = -9, C = 4$)

$\delta = B^{2} - 4\cdot A\cdot C$

$\delta = (-9)^{2}-4\cdot (12)\cdot (4)$

$\delta = -15$

$y = 12\cdot x^{2}-9\cdot x + 4$ has no real solutions.

2) $10\cdot x + y = -x^{2}+2$

First, we reorganize the expression:

$y = -x^{2} -10\cdot x + 2$ ($A = -1, B = -10, C = 2$)

$\delta = B^{2} - 4\cdot A\cdot C$

$\delta = (-10)^{2}-4\cdot (-1)\cdot (2)$

$\delta = 108$

$10\cdot x + y = -x^{2}+2$  has two distinct real solutions.

3) $4\cdot y - 7 = 5\cdot x^{2}-x + 2 + 3\cdot y$

First, we simplify the expression:

$4\cdot y - 3\cdot y = 5\cdot x^{2}-x +2 +7$

$y = 5\cdot x^{2}-x +9$ ($A = 5$, $B = -1$, $C = 9$)

$\delta = B^{2} - 4\cdot A\cdot C$

$\delta = (-1)^{2}-4\cdot (5)\cdot (9)$

$\delta = -179$

$4\cdot y - 7 = 5\cdot x^{2}-x + 2 + 3\cdot y$ has no real solutions.

4) $y = (-x +4)^{2}$

First, we expand the expression:

$y = x^{2}-8\cdot x + 16$ ($A = 1$, $B = -8$, $C = 16$)

$\delta = B^{2} - 4\cdot A\cdot C$

$\delta = (-8)^{2}-4\cdot (1)\cdot (16)$

$\delta = 0$

$y = (-x +4)^{2}$ has two equal real solutions.

We kindly invite to check this question on polynomials: https://brainly.com/question/15465256