Formulaic Maths Problems - get another problem: level 1, level 2, level 3 - link to this problem
Loading, please wait...
Find all real solutions to $-5x^2-8x+ 8 = 0$ using the quadratic formula.
Find the discriminant. Hint: The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $\Delta = b^2 - 4ac$.
$\Delta = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~} {}^2 - 4 \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{54px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{54px}}~}$ $\Delta = -8 {}^2 - 4 \times -5 \times 8$
$\Delta = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{54px}}~}$ $\Delta = 64 - -160 = 224$
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 $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$.
$x = \dfrac{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} \pm \sqrt{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~}}}{2 \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C3}}\hspace{54px}}~}}$ $x = \dfrac{8 \pm \sqrt{224}}{2 \times -5}$
Hint: $x = \dfrac{-b + \sqrt{\Delta}}{2a}$ or $x = \dfrac{-b - \sqrt{\Delta}}{2a}$.
$x = \dfrac{8 \pm \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~}}{\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~}}$ $x = \dfrac{8 \pm 14.96662955}{-10}$
Therefore,
$x = \dfrac{\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~}}{-10} \text{ or } x = \dfrac{\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~}}{-10}$ $x = \dfrac{22.96662955}{-10} \text{ or } x = \dfrac{-6.96662955}{-10}$
$x = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \text{ or } x = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~}$ $x = -2.29666295 \text{ or } x = 0.69666295$