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Find all real solutions to $x^2+ x-9 = 0$ using the quadratic formula.
Find the discriminant. Hint: The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $\Delta = b^2 - 4ac$.
$\Delta = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~} {}^2 - 4 \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{54px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{54px}}~}$ $\Delta = [1] {}^2 - 4 \times [1] \times [-9]$
$\Delta = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{54px}}~}$ $\Delta = [1] - [-36] = [37]$
Hint: If $\Delta > 0$ there are two real solutions, if $\Delta = 0$ there is only one, and if $\Delta < 0$ there are none.
So the number of real solutions is ${\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{noOfSolutions}}\hspace{54px}}~}}$. So the number of real solutions is ${[2]}$.
Substitute into the formula. Hint: The quadratic formula states $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$.
$x = \dfrac{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} \pm \sqrt{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~}}}{2 \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C3}}\hspace{54px}}~}}$ $x = \dfrac{[-1] \pm \sqrt{[37]}}{2 \times [1]}$
Hint: $x = \dfrac{-b + \sqrt{\Delta}}{2a}$ or $x = \dfrac{-b - \sqrt{\Delta}}{2a}$.
$x = \dfrac{-1 \pm \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~}}{\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~}}$ $x = \dfrac{-1 \pm [6.08276253]}{[2]}$
Therefore,
$x = \dfrac{\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~}}{2} \text{ or } x = \dfrac{\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~}}{2}$ $x = \dfrac{[5.08276253]}{2} \text{ or } x = \dfrac{[-7.08276253]}{2}$
$x = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \text{ or } x = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~}$ $x = [2.54138127] \text{ or } x = [-3.54138127]$