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Formula
Solve the quadratic equation
Answer
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Number of solution
Answer
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Relationship between roots and coefficients
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Graph
$y = x ^ { 2 } + 3 x - 70$
$y = 0$
$x$-intercept
$\left ( 7 , 0 \right )$, $\left ( - 10 , 0 \right )$
$y$-intercept
$\left ( 0 , - 70 \right )$
Minimum
$\left ( - \dfrac { 3 } { 2 } , - \dfrac { 289 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } - \dfrac { 289 } { 4 }$
$x ^{ 2 } +3x-70 = 0$
$\begin{array} {l} x = 7 \\ x = - 10 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 70 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 10 } \end{array}$
$\begin{array} {l} x = 7 \\ x = - 10 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 70 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 70 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ 70 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = 70 + \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = 70 + \dfrac { \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 70 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 ^ { 2 } } { 2 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 289 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 289 } { 4 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 289 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 289 } { 4 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 17 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 17 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 17 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 17 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 17 } { 2 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 10 } \end{array}$
$\begin{array} {l} x = 7 \\ x = - 10 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 1 \times \left ( - 70 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 289 } } { 2 \times 1 } }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 289 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 3 \pm \color{#FF6800}{ 17 } } { 2 \times 1 }$
$x = \dfrac { - 3 \pm 17 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 3 \pm 17 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm 17 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 + 17 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 - 17 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 17 } } { 2 } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$ $ Add $ - 3 $ and $ 17$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 14 } } { 2 } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 14 } { 2 } } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 7 } { 1 } } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 7 } { 1 } } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 7 } \\ x = \dfrac { - 3 - 17 } { 2 } \end{array}$
$\begin{array} {l} x = 7 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 17 } } { 2 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 7 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 20 } } { 2 } \end{array}$
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ \dfrac { - 20 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ \dfrac { - 10 } { 1 } } \end{array}$
$\begin{array} {l} x = 7 \\ x = \dfrac { - 10 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 10 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 70 } \right )$
$D = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 70 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 9 } - 4 \times 1 \times \left ( - 70 \right )$
$D = 9 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 70 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 9 - 4 \times \left ( - 70 \right )$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 70 } \right )$
$ $ Multiply $ - 4 $ and $ - 70$
$D = 9 + \color{#FF6800}{ 280 }$
$D = \color{#FF6800}{ 9 } \color{#FF6800}{ + } \color{#FF6800}{ 280 }$
$ $ Add $ 9 $ and $ 280$
$D = \color{#FF6800}{ 289 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 289 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 3 , \alpha \beta = - 70$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ In the quadratic equation $ ax^{2}+bx+c=0 $ , if the two roots are $ \alpha, \beta $ , then it is $ \alpha + \beta =-\dfrac{b}{a} $ , $ \alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 70 } { 1 } }$
$\alpha + \beta = - \dfrac { 3 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 70 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 3 } , \alpha \beta = \dfrac { - 70 } { 1 }$
$\alpha + \beta = - 3 , \alpha \beta = \dfrac { - 70 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 3 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 70 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-III. Factoring Polynomials 
Find the factors of the following 
polynomials. 
Examples; 
$1$ $x^{2}-36=\left(x+6\right)\left(x-6\right)$ 
$2$ $x^{2}-x-6=\left(x-3\right)\left(x+2\right)$ 
$3$ $2x^{2}-4x=2x\left(x-2\right)$ 
$1.x^{2}-25=$ 
$2.x^{2}-100=$ 
$4x^{2}-36=$ 
$3.4x^{2}$ 
$4.5x^{3}-10x=$ 
$5.15x-3=$ 
$6.27x^{4}+9x^{2}=$ 
$7.3x^{2}+19x-14=$ 
$8.x^{2}+3x-70=$ 
$9.30x^{2}-7x-2=$ 
$10.$ $4x^{2}+17x-15=$
1st-6th grade
Geometry
search-thumbnail-$\left(x-5\right)m$ $\left(x-2\right)m$ 
$\left(x-5\right)m$ 
$D$ $C$ 
Write down an expression in terms of $x$ for the area of the field $ABCD$ 
(а) Expand and simplify your answer. 
As shown in the diagram above, a square with length $\left(x-5\right)m$ inside the field is fenced up to 
plant vegetables. 
Write down an expression in $tems-of$ $x$ for the area of the square. 
(b) $\right)$ Expand and simplify your answer. 
$\left($ (c) Given that the shaded area of the rectangular field is $51m^{2}$ , write down an equation in $x$ 
and show that it reduces to $x^{2}+3x-70=0$ 
(d) Solve the equation $x^{2}+3x-70=0$ 
$\left($ (e) Find the area of the rectangle field $ABCD$
10th-13th grade
Other
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