Algebra 9.5 Solving Quadratic Equations Using the Quadratic Formula This is an important section as there are many questions on the STAR test about the.

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Algebra 9.5 Solving Quadratic Equations Using the Quadratic Formula This is an important section as there are many questions on the STAR test about the quadratic formula.

Solve 2x = x 2 = 18 x 2 = 9 x = + 3 These are the solutions/roots of the equation. We did not need the quadratic formula to solve this quadratic equation because it was in the form… Ax 2 + C = # where b = 0.

What is the quadratic formula? It is a formula used to solve any quadratic equation in the form… ax 2 + bx + c = 0 when a ≠ 0 and b 2 – 4ac ≥ 0. Using the formula will produce the solutions(roots) of the equation.

Here it is…try to memorize it… x = This formula can be used to find the roots of any quadratic equation in the form ax 2 + bx + c = 0. b 2 – 4ac 2a -b +

Find the roots of 2x = 28 using the quadratic formula… The equation must be in the form ax 2 + bx + c = 0 before using the quadratic formula. 2x = x = 0 Remember the roots were x = + 3 Must be 0 in order to use the quadratic formula. -b + b 2 – 4ac 2a X = -0 + (0) 2 – 4(2)(-18) 2(2) X = X = X = 3 and - 3 X = a = 2, b = 0, c = -18 These are the roots of the equation.

Find the roots of -3x 2 + 4x = -5 using the quadratic formula… -3x 2 + 4x = x 2 + 4x + 5 = 0 -b + b 2 – 4ac 2a X = -4 + (4) 2 – 4(-3)(5) 2(-3) X = a = -3, b = 4, c = (-3) X = X = X = Can you reduce? Both numbers in the numerator must have common factors of the denominator X = These are the two roots. and Yes, by -2.

You try! Find the roots of 4x 2 - x = 7 using the quadratic formula… who can do it on the board? 4x 2 - x = x 2 - x - 7 = 0 -b + b 2 – 4ac 2a X = 1 + (-1) 2 – 4(4)(-7) 2(4) X = a = 4, b = -1, c = X = X = These are the two roots of the equation and Can you reduce? No.

Note: This is on the STAR test. The roots of a quadratic equation are the x-intercepts of the graph of the quadratic (parabola). The roots = x-intercepts

You try! Find the x-intercepts of the graph of y = x 2 + 5x – 6. using the quadratic formula… who can do it on the board? y = x 2 + 5x – 6 0 = x 2 + 5x – 6 (substitute 0 for y) -b + b 2 – 4ac 2a X = -5 + (5) 2 – 4(1)(-6) 2(1) X = a = 1, b = 5, c = X = X = These are the two roots and x-intercepts and 2 2 X = and 1 X = -6 and

One from the HW P. 536 #46

HW P # 33-36, 42-46, Leave answers in simplified radical form.