Respuesta :
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
[tex]y = 12x^2 - 9x+ 4[/tex]
where, a = 12, b = -9 and c = 4
x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]
x = [tex]-(-9)+ \sqrt{(-9)^2 - 4(12)(4)} /2(12)[/tex]
x = 1.063 or -0.313
Second Equation
[tex]10x +y = -x^2 +2 \\y = x^{2} +10x - 2[/tex]
where, a = 1, b = 10, c = -2
substitute the values into the equation
x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]
[tex]-10 + \sqrt{10^2 - 4(1)(-2)} /2(1)[/tex]
x = 0.195 or -10.195
Third Equation
[tex]4y - 7 = 5x^2 - x +2 +3y[/tex]
[tex]y = 5x^{2} -x -5[/tex]
where, a = 5, b = -1, c = - 5
x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]
[tex]-(-1) + \sqrt{(-1)^2 - 4(5)(-5)} /2(5)[/tex]
x = 0.905 or -0.55
Fourth Equation
[tex]y = (-x +4)^2 \\y = x^{2} -8x +16[/tex]
where, a = 1, b = -8, c = 16
x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]
x = [tex]-(-8) + \sqrt{(-8)^2 - 4(1)(16)} /2(1)[/tex]
x = 4
Learn more about quadratic equations;
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