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

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

What is a quadratic equation?

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

The given quadratic equation

$y = 12x^2 - 9x+ 4$

where, a = 12, b = -9 and c = 4

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

x = $-(-9)+ \sqrt{(-9)^2 - 4(12)(4)} /2(12)$

x = 1.063 or -0.313

Second Equation

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

where, a = 1, b = 10, c = -2

substitute the values into the equation

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

$-10 + \sqrt{10^2 - 4(1)(-2)} /2(1)$

x = 0.195 or -10.195

Third Equation

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

$y = 5x^{2} -x -5$

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

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

$-(-1) + \sqrt{(-1)^2 - 4(5)(-5)} /2(5)$

x = 0.905 or -0.55

Fourth Equation

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

where, a = 1, b = -8, c = 16

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

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

x = 4

Learn more about quadratic equations;

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