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Formula
Solve the quadratic equation
Answer
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Number of solution
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Relationship between roots and coefficients
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Graph
$y = \left ( x + 4 \right ) ^ { 2 } - 25$
$y = 0$
$x$-intercept
$\left ( 1 , 0 \right )$, $\left ( - 9 , 0 \right )$
$y$-intercept
$\left ( 0 , - 9 \right )$
Minimum
$\left ( - 4 , - 25 \right )$
Standard form
$y = \left ( x + 4 \right ) ^ { 2 } - 25$
$\left( x+4 \right) ^{ 2 } -25 = 0$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
Find solution by method of factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Expand the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
Solve quadratic equations using the square root
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
Calculate using the quadratic formula
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm \sqrt{ 100 } } { 2 \times 1 } }$
$x = \dfrac { - 8 \pm \sqrt{ \color{#FF6800}{ 100 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 8 \pm \color{#FF6800}{ 10 } } { 2 \times 1 }$
$x = \dfrac { - 8 \pm 10 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 8 \pm 10 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm 10 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 + 10 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 - 10 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$ $ Add $ - 8 $ and $ 10$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 2 } } { 2 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 2 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 1 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 2 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 18 } } { 2 } \end{array}$
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ \dfrac { - 18 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ \dfrac { - 9 } { 1 } } \end{array}$
$\begin{array} {l} x = 1 \\ x = \dfrac { - 9 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$D = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 9 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 64 } - 4 \times 1 \times \left ( - 9 \right )$
$D = 64 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 9 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 64 - 4 \times \left ( - 9 \right )$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$ $ Multiply $ - 4 $ and $ - 9$
$D = 64 + \color{#FF6800}{ 36 }$
$D = \color{#FF6800}{ 64 } \color{#FF6800}{ + } \color{#FF6800}{ 36 }$
$ $ Add $ 64 $ and $ 36$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 8 , \alpha \beta = - 9$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 8 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 9 } { 1 } }$
$\alpha + \beta = - \dfrac { 8 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 9 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 8 } , \alpha \beta = \dfrac { - 9 } { 1 }$
$\alpha + \beta = - 8 , \alpha \beta = \dfrac { - 9 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 8 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 9 }$
$ $ 그래프 보기 $ $
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7th-9th grade
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search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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10th-13th grade
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