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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
$y = 4 x ^ { 2 } + 10 x + 6$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 3 } { 2 } , 0 \right )$, $\left ( - 1 , 0 \right )$
$y$Intercept
$\left ( 0 , 6 \right )$
Minimum
$\left ( - \dfrac { 5 } { 4 } , - \dfrac { 1 } { 4 } \right )$
Standard form
$y = 4 \left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } - \dfrac { 1 } { 4 }$
$4x ^{ 2 } +10x+6 = 0$
$\begin{array} {l} x = - 1 \\ x = - \dfrac { 3 } { 2 } \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$\begin{array} {l} x = - 1 \\ x = - \dfrac { 3 } { 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } = - \dfrac { 3 } { 2 } + \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } = - \dfrac { 3 } { 2 } + \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 ^ { 2 } } { 4 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 1 } { 16 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 1 } { 16 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 1 } { 16 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 1 } { 16 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$\begin{array} {l} x = - 1 \\ x = - \dfrac { 3 } { 2 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = 0$
$ $ Bind the expressions with the common factor $ 2$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 0 }$
$ $ Divide both sides by $ 2$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 2 \times 3 } } { 2 \times 2 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 \pm \sqrt{ 1 } } { 2 \times 2 } }$
$x = \dfrac { - 5 \pm \sqrt{ \color{#FF6800}{ 1 } } } { 2 \times 2 }$
$n $ square root of 1 is 1 $ $
$x = \dfrac { - 5 \pm \color{#FF6800}{ 1 } } { 2 \times 2 }$
$x = \dfrac { - 5 \pm 1 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Multiply $ 2 $ and $ 2$
$x = \dfrac { - 5 \pm 1 } { \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 \pm 1 } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 + 1 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 - 1 } { 4 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } { 4 } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$ $ Add $ - 5 $ and $ 1$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 4 } } { 4 } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 4 } { 4 } } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 1 } { 1 } } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$\begin{array} {l} x = \dfrac { - 1 } { \color{#FF6800}{ 1 } } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ x = \dfrac { - 5 - 1 } { 4 } \end{array}$
$\begin{array} {l} x = - 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 4 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = - 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } } { 4 } \end{array}$
$\begin{array} {l} x = - 1 \\ x = \color{#FF6800}{ \dfrac { - 6 } { 4 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = - 1 \\ x = \color{#FF6800}{ \dfrac { - 3 } { 2 } } \end{array}$
$\begin{array} {l} x = - 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 2 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = - 1 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 10 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$
$D = \color{#FF6800}{ 10 } ^ { \color{#FF6800}{ 2 } } - 4 \times 4 \times 6$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 100 } - 4 \times 4 \times 6$
$D = 100 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$
$ $ Multiply the numbers $ $
$D = 100 \color{#FF6800}{ - } \color{#FF6800}{ 96 }$
$D = \color{#FF6800}{ 100 } \color{#FF6800}{ - } \color{#FF6800}{ 96 }$
$ $ Subtract $ 96 $ from $ 100$
$D = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 4 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = \dfrac { 3 } { 2 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 4 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 6 } { 4 } }$
$\alpha + \beta = - \color{#FF6800}{ \dfrac { 10 } { 4 } } , \alpha \beta = \dfrac { 6 } { 4 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - \color{#FF6800}{ \dfrac { 5 } { 2 } } , \alpha \beta = \dfrac { 6 } { 4 }$
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = \color{#FF6800}{ \dfrac { 6 } { 4 } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-II. Solve the following in a Clean Sheet of Paper 
$A$ Determine the sum and product of the roots of 
each equation. 
$1.$ $x^{2}+7x+10=0$ 
2. $x^{2}-5x+6=0$ 
$3.$ $x^{2}-12x=0$ 
$4$ $x^{2}+16=0$ 
5. $x^{2}+3x-5=0$ 
$6.$ $x^{2}-7x+12=0$ 
$7.$ $x^{2}+4x+4=0$ 
$8.$ $x^{2}+10x+25=0$ 
$9.$ $x^{2}+12x+20=0$ 
$10.x^{2}+2x-15=0$
7th-9th grade
Other
search-thumbnail-2 Solve each of the following equations by using the quadratic formula. Give your answers 
correct to two decimal places where necessary. These quadratic expressions do not factorise. 
a $x^{2}+6x+4=0$ b x+x-4=0 $x^{2}+14x+44=0$ 
d $x^{2}-6x-8=0$ e $x^{2}+10x+17=0$ f $x^{2}+7x+1=0$ 
g $x^{2}+24x+121=0$ h $x^{2}+11x+27=0$ i $x^{2}+18x-93=0$ 
$x^{2}-2x-10=0$ k $x^{2}-8x-5=0$ 1 $x^{2}-2x-100=0$ 
3 Solve each of the following equations. Give your answers correct to two decimal places 
where necessary. 
a $x^{2}-4x-8=0$ b $6x^{2}+11x-35=0$ c $2x^{2}-3x+\dfrac {1} {2}=0$ 
d $2x^{2}-5x=25$ e $4x^{2}-13x+9=0$ f 8x-4x =18 
9x-136x $h$ $-x\left(x-16\right)+57=0$ i 9(x+1)= x 
g j 2x +5x 3 k $6x^{2}+13x+6=0$ 4x -x-3=0
7th-9th grade
Algebra
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