When you would not need to use the quadratic formula

1 answer:

3 0

Answer: When the quadratic equation can be factored or solved by completing the square, you do not need to use the quadratic formula

Step-by-step explanation: In the form ax² + bx + c

Whenever a is 1 (implied, no coefficient of x²) and the constant, c, is a multiple of numbers whose sum or difference is equal to b, factoring is much quicker and easier than using the quadratic formula.

Even when there is a coefficient of x², there are procedures to find the factors and solve for x without resorting to the quadratic formula.

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Solve the equation using the quadratic formula : x^2-10x+25=18

Sever21 [200]

$x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}$

$Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}$

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3 years ago

Find the indicated values, where g(t)=t^2-t and f(x)=1+x f(2g(1)

Darina [25.2K]

$g(t)=t^2-t\\\\f(x)=1+x\\\\f(2g(1))=?\\\\g(1)=\\_{\text{put t=1 to the equation of g(t)}}\\\\=1^2-1=1-1=0\\---------------\\2g(1)=2(0)=0\\\\f(2g(1))=\\_{\text{put x-0 to the equation of f(x)}}\\\\=1+0=1\\\\Answer:\ f(2g(1))=1$