Question

Determine the number of real solutions each quadratic equation has.
y = 12x2 - 9x + 4 real solution(s)
10x + y = -x2 + 2 real solution(s)
4y - 7 = 5x2 - x + 2 + 3y real solution(s)
y = (-x + 4)2 real solution(s)

Answer 1

Answer: no , two , no and one

Step-by-step explanation:

Answer 2

All the equations have 2 real roots or solutions except for equation 4 which have only one real root.

To find the number of real solutions the quadratic equation have, we would use quadratic formula to do this.

$x = \frac{-b +- \sqrt{b^2 -4ac} }{2a}$

First Equation,

data given;

• a = 12
• b = -9
• c = 4

substitute it into the formula

$\frac{-(-9)+-\sqrt{(-9)^2 - 4(12)(4)} }{2(12)}\\\frac{9 +- 16.52 }{24}\\x = 1.063 or x = -0.313$

Second Equation

data given

$10x+y = -x^2 +2\\y = -x^2-10x+2\\y = x^2 +10x-2$

• a = 1
• b = 10
• c = -2

substitute the values into the equation

$\frac{-b +- \sqrt{b^2 -4ac } }{2a} \\\frac{-10+- \sqrt{10^2 -4*1*-2} }{2*1} \\x = 0.195 or x = -10.195$

Third Equation

data given;

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

• a = 5
• b = -1
• c = - 5

$\frac{-(-1)+- \sqrt{(-1)^2-4(5)(-5)} }{2(5)} \\x = 0.905 or x = - 0.55$

Fourth Equation

$y = (-x + 4)^2\\y = x^2 +2(-4x)+16\\y = x^2 -8x + 16$

• a = 1
• b = -8
• c = 16

substitute into the equation and solve

$\frac{-b+-\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-(-8)+- \sqrt{(-8)^2 -4(1)(16)} }{2(1)}\\x = 8/2 \\x = 4$

from the calculations above, all the equation have two real roots except for the last one which have only one real root.

Learn more about quadratic roots here;

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