Use any method to solve the following equations, giving your answers to 2 decimal places where necessary. If an equation has no real roots, state this is so.
1. x2-5x+7=0
2. (3x-4)=(4x-3)2
1. This is already in standard form. You should be able solve that equation from what you've learned.
2. Here it is convenient to substitute.
y = 4x-3
Then
x = 1/4(y+3) -------->
3x - 4 = 3/4 (y+3) -4 =
3/4 y - 7/4
So, the equation is:
3/4 y - 7/4 = y^2 -------->
4 y^2 - 3 y + 7= 0
To solve the given equations, we will use different methods depending on the type of equation. Let's solve each equation step by step:
1. x^2 - 5x + 7 = 0
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac))/(2a)
Comparing the given equation with the standard form ax^2 + bx + c = 0, we have a = 1, b = -5, and c = 7. Plugging these values into the quadratic formula:
x = (-(-5) ± √((-5)^2 - 4(1)(7)))/(2(1))
x = (5 ± √(25 - 28))/2
x = (5 ± √(-3))/2
The expression √(-3) indicates the square root of a negative number, which is not a real number. Therefore, the equation x^2 - 5x + 7 = 0 has no real roots.
2. (3x - 4) = (4x - 3)^2
Simplifying the equation, we have:
3x - 4 = (4x - 3)(4x - 3)
3x - 4 = 16x^2 - 24x + 9
Let's rearrange the equation to form a quadratic equation:
16x^2 - 24x - 3x - 4 - 9 = 0
16x^2 - 27x - 13 = 0
This is a quadratic equation. We can again solve it using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
Comparing the given equation with the standard form ax^2 + bx + c = 0, we have a = 16, b = -27, and c = -13. Plugging these values into the quadratic formula:
x = (-(-27) ± √((-27)^2 - 4(16)(-13)))/(2(16))
x = (27 ± √(729 + 832))/32
x = (27 ± √1561)/32
These values cannot be simplified further. Therefore, the solutions to the equation (3x - 4) = (4x - 3)^2 are:
x = (27 + √1561)/32 and x = (27 - √1561)/32, rounded to 2 decimal places if necessary.